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BIBLIOGRAPHIC RECORD TARGET 

Graduate Library 
University of Michigan 

Preservation Office 

Storage Number: 



ACV0278 

UL FMT B RT a BL m T/C DT 07/19/88 R/DT 07/19/88 CC STATmmE/Ll 
035/1: : | a (RLIN)MIUG86-B35983 
035/2: : j a (CaOTULAS)l 60645772 
040; : | a MiU | c MiU 
100:1 : I a Durell, Fletcher, | d 1859- 

245:00: | a Plane and solid geometry, | c by Fletcher Durell. 
260: : | a New York, | b Charles E. Merrill co. (c 1911 [cl904] 
300/1: : |avi, 7-552p. | b front, (group of ports.) diagrs. |cl9cm. 
490/1:0 : | a Durell's Madiematical series 
650/1:0: | a Geometry 
998: : |cWFA |s9124 



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On behalf of 

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PLANE AND SOLI! 
GEOMETRY 

BY 

FLETCHER DUEEI.L, Pii.D. 



NEW YORK 
CHARLES E. MERRILL CO. 

44-60 East Twentv-thtrd Street 
1911 



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DuroiPs r^Iathf 



Plane Geometry 

341 pages, 12mo, hiiii Icatlu-r 

Solid Geometry 

213 pages, 12nio, lialf leatliev 

l^lane and Solid Geometry 

614 pages, 12ino, half leather 

I'lane Trigonometry 

184 pages, Svo, oloth . . . 

Plane Trigonometry and Tallies 
298 pages, Svo, uloth . . . 



P|ane Tiigonometry, with Surveying 
Tables 
In preparation 



Plane and Spherical Trigonometvy, 
Tables 

351 pages, Svo, cloth .... 



Plane and Spherical Trigonometry, 
Surveying and Tahles 
In preparation 



Logarithmic and Trigonometric Table; 
114 pages, Svo, cloth .... 



Copyright, 101)4, by Cliarles E. Merrill Co. 
[5] 



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PREFACE 

One of tlie main pui-poses in wi'itiug this book has 
been to try to present the subject of Geometry so tliat llie 
pupil shall Tiuderatand it not merely as a series of eorrei^t 
deductions, but shall realize the value and meaning of its 
principles as well, This lapect of tlie subject has been 
directly presented in some places, and it is hoped that it per- 
vades and shapes the presentation in all places. 

Again, teachers of Geometry generally agree that the 
most difficult part of their work lies in developing in 
pupils the power to work original eseixsises. The second 
main purpose of the book is to aid in the solution of this 
difiieulty by arranging original exercises iu groups, each 
of the earlier groups to be worked by a distinct method. 
The pupil is to be kept woi-king at each of these groups 
till he masters the method involved in it. Later, groups 
of mixed exercises to be worked by various methods are 
given. 

In tlie current exercises at the bottom of the page, 

only such exercises are used as can readily be solved in 

connection with the daily work. All difflcalt originals are 

included in the groups of exercises as indicated above. 

yimilarly, iu the writer's opinion, many of the numeri- 

m 



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cal applications of geometry call for special metliods of 
solution, and the thorough treatment of such exercises 
should be taken up separately and systematic ally, [See 
pp. 304-318, etc.] In the daily extempore work only such 
numerical problems are included as are needed to make 
clear and definite the meaning and value of the geometric 
principles considered. 

Every attempt has been made to create and cultivate 
the heiiristic attitiiile on the part of the pupil. This has 
been done by the method of initiating the pupil into 
original work described above, by qiieries in the course 
of proofs, and also at the bottom of different pages, and 
also by occasional queries in the course of the text where 
definitions and discussions are presented. In the writer's 
opinion, the time has not yet come for the purely heuristio 
study of Geometry in most schools, but it is all-important to 
use every means to arouse in the pnpil the attitude and 
energy of original investigation in the study of the siihject. 

In other respects, the aim has been to depart as little 
as possible from the methods most generally used at 
present in teaching Keometi-y. 

The Practical Applications (Groups 88-91) have been 
drawn from many sources, but the autliur wishes to ox- 
prcKs his especial indebtedness to the Committee which has 
collected the Real Applied Problems published from time 
to time in Sfhool Science nnd Miitheinatii.-x^ and of which 
Professor J. F. ilillis of the Francis "W. Parker School 
of Chicago is the chairman. Page 360 is dne almost en- 
tirely to Professor AViiHam Eetz of the East High School 
of Koehester, N. Y. 

FLETCHER DURELL. 
LiwaBNCEyiLi.E, N. J., Sept. 1, 1904, 



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TO THE TBACHEK 

1. In working original exercises, one of the chief dif- 
ficulties of pupils lies ill their inability to construct the 
figure required and to make the particular enunciation 
from it. Many pupils, who are quite unable to do this 
preliminary work, after it is done can readils' discover a 
proof or a solution. In many csereisea in this book the 
figure is drawn and the particular enunciation made. It 
is left to the discretion of the teacher to determine for 
what other exercises it is best to do this for pupils. 

2. It is frequently important to give partial aid to the 
pupil by eliciting the outline of a proof by questions such 
as the following: "On this figure (or, iu these two tri- 
angles) what angles are equal, and why?" "What lines 
are equal, and why?" etc. 

3. In many cases it is also helpful to mark iu colored 
crayon paii's of equal lines, or oE equal angles. Thus, in 
the figure on p. 37 lines AB and DE may be drawn with 
red crayon, AO and DF with blue, and the angles A and J> 
marked by small arcs drawn with green crayon. IE 
colored craj"ons are not at hand, the homologous equal 
parts may be denoted by like synibois placed on them, 

tllUS : F 




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^ TO THE TE.U'HER 

(either given or to be proved) by lifi- 
ures denoting the onk^r in which 11.,/ 
lines are to be taken. Thus, if 0.1: 
0C= 01)  OB, the relation iimy !)e 
indicated as ou the figure. 

4. It is sometimes helpful to vaiy 
the symbolism of the book. Thus, in dealing 
equalities a convenient symbol for "angle" is 2)^, 

aa ^ A > 2^1 B. 

5. Bach pupil need be required to work only so many 
originals iu each group as will give him a mastery of the 
particular method involved. A large number of exercises 
is given in order that the teacher may have mnuy to select 
from and may vary the work with successive classes. 

6. It is important to insist that the solutions of exer- 
cises for the first few weeks be carefully written out; later, 
for many pupils, oral demonstration will be suiilcieiit.and 
ground can be covered more rapidly "^y its use. 

7. In leading pupils to appreciate the meaning of 
theorems, it is helpful at times to point out that not every 
theorem has for its oljject the demonstration of a new and 
unexpected truth (i, e., not all are "synthetic"), but that 
some theoreips are analytic, it being their purpose to re- 
duce an obvious truth to the certain few principles with 
which we start in Geometry. Their function is, therefore, 
to simplify and clarify the subject rather tbau to extend 
its content. 



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TABLE OF CONTENTS 



Definitions akd Fiiwr Piunciplks . . 
Book I. Rectilinear Fh;uhes .... 
Boos II. The Cihclb 

E(.aK III. PKOPORTiaS. SIJKI.AU PoLVt 

Book IV. Aki:a i-i' PoLvi;(.Na , . , . 
Book V. Eeri-l.^r Polyiions. Mkahurf 
KuiiDiutjAL Applications or I'lanl V.\:f 
Book VI. Likes, Planes and Anolks 

BofJK Vil. l'OLVHKI)liO\8 

Booii VIH, CVE,INDK1:S A.SD C'o.NLiSi . 

Book IX. The Si'iiiiitE 

MojiEincAi. Esn;RC(SEs in Somd Geomi 
MoDEiis (JEOJfETiiic Concepts .... 

IIlSTOHV OF UHOMl-.THV 

Review E.'iEiicisi'is in Pi.ani; tlr.oMTyn 
Review E,\-ehcisi;s in Solid (iEoMEii: 
I'R.vrncAL Api'i,icatio\s of Ceomf.ti 



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SYMBOLS AND ABBREVIATIONS 



+ plus, or increased hg. 


Adj. . . 


adjareiit. 


— minus, or dimimshed hj. 


Alt. . 


alternate. 


X multiplied hj. 


Art. . 


article. 


-i- divided hy. 


Ax,. . 


ax>«m. 


~ equals; is (or arc) equal lo. 


Coiistr. 


conslrucliox.. 


A approaches {as a liiiiit] . 


Cor, . 


corollary. 


K= is (or are) equivalent to. 


Def. . 


dcfimtion. 


> is (or ace) greater than. 


Ex.. . 


exercise. 


< is {or are} less than. 


Est. . 


exterior. 


.: there/ere. 


Fig. . 


fignrc. 


X perpejidiCKlar , pcrpeiKUciihir t 


, Hyp. . 


hypothesis. 


or, is perpendicular Id. 


Ident. 


identity. 




Ii.t. . 


interior. 


it parallel, or, is parallel lo. 


I'oBt. . 




Ila parallels. 


Prop. . 


. propositioH. 


I , i angle, angles 


m. . . 


right. 


A, A triangle, triangles. 


Sns. . 


. suggestion. 


CJ , E£! parallelogran}, paralklogram.i 


Sup. . 


supplemciila 


O, ® circle, circles. 


St. . . 


. sfmigbt. 


q. E. D, quod erat demonsfrandum 


that is, w 


hich Wiis 10 


proved. 






Q. E. F. quod erat faeiendum; that i 


, which was 


to be ma^ie. 



A few other abbreriatione and symbols will be introduced i 
their meaning iodicaied latei: on. 



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DEFINITIONS AND FIRST PRINCIPLES 

IKTRODOCTORY ILLUSTRATIONS. DEFIHITIOWS 

1, Computation of an area, Pr;ieticiil experience lias 
taught men that cei'fcaiii ways of dealing with objects in 
the world about us are more advuiifageons tlmn others; 
thus, if it be desired to find the num- 
ber of square yards in the area of a 
floor, we do not mark off the floor 
iuto actual square yards and count 
the number of square yards thus 
made, but pursue the much easier 
course of measuring two lines, the 
length and breadth of the floor, obtaiuicg 7 yards, say, as 
the length, and 5 yards as the width, and multiplying 
the length by the breadth. The area is thus found to be 
35 squai'e yards. 

Let the pupil determine the area o£ soioe couveuient floor iu oauli 
of these two waj-s, aud compare the inbof of the two processes. 

2. Computation of a volume. Similarly, for example, 
in order to determine the n amber of cubic feet which a 
box contains, instead of filling the box with block.s of 
■Wood, each of the size of a eubie foot, and 
counting the number of blocks, we pur- 
sue the much easier course of measuring 
the number of linear feet in each inside 
edge of the box and multiplying together 
the three dimensions obtained; thus, if 

19) 




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]0 PLANE GROHrETE"/ 

tlie inside dimonMioii,-: an: », ■! anil 3 fci-'t, llif volumu is 
5 X 4 X 3, or CO cubic tuat. 

Similarly, the direet methoii of mensuriiif! the nuniTjer ii£ bushels 
oE wheat in a bin is to fill a bushel meaaure with wheat from thu 
bin, time after time, tiH the biu is exhausted, and count the number 
of times the bushel measure is used. But a much leas laborious 
method is to measure the three dimensions of the bin in inches 
and divide their pvoduet by the number of eubic inches in a buabel. 

Let tlie pupil in like manner compare the lal)or of finding the 
number of feet of lumber in a given bloek of wood by actually saw- 
ing the Mock np into lumber feet, with the labor of measni'lng the 
dimensions of the block a7id computing the number of lumber feet 
by taking the product of the linear dimensions obtained. 

3. Unknown line determined from known lines. Tlie 
fituiloLit is also probably familiar with the faut tluit, ii.v 
comjiutiitionB based on the relations of cer- 
tain lines whose lengths are known, the 
lengths of other unknown lines may be 
determined without the labor of measuring 
these unknown lines. 




Thus, for instance, if a ladder 5 yards long lean 
agninst a wall and have ita foot 3 yards from the 
wall, the hoight of the top of the ladder from the 
ground may be determined thus: 

(Height)^ = 5'^ _ 3' = 23 — 9 = IG. 

4. Economies in representing surfaces, lines, etc. Other 
principles of advantage of an even more general eharaeter 
occur in dealing with geometric objects. Thus, since only 
one straight line can be passed through two given points, 
tlie two points may be taken as a highly economized 
symbol or representative of the line, by the use of which 
much labor is saved in dealing with lines, and new results 
are made attainable. 



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riT.nMV.TIilf JtAnNITUDES 11 

Similarly, siiieo only one fliit, or plane surf:iee enn be 
passed through three given points (not iu the same straight 
line), these three points may be taken as a symbol or 
representative of the tlat sni-face. This gives the advan- 
tage not only of reducing an nnlimited surface to three 
points, but also of giving for the plane a symbol made up 
of three parts. By varying one of these parts and not the 
others, tlie plane iriiiy be varied in ouo respeet and not in 
others; also planes having certain properties in <;onimon 
may he grouped togetlier, and dealt with iu the groups 
formed. 

5. Geometry as a science. Definition. The above illus- 
trations serve to Amw limv advantageous it often is to deal 
with geometric magnitudes by certain methods rather than 
others. 

In the study of Geometry as a science we proceed to 
niake a systematic examination of these methods. 

Geometry is the science which treats of the properties of 
continuous magnitudes and of space. 

GEOMETRIC MAGNITUDES 

6. Solids. A physical solid is a portion of matter, as a 
bioek of wood, an iron weight, or a piece of marble. 

The portion of space oceiipied by a physieal solid may 
be considered apart from the physieal solid itself, hence 

A geometric solid is the portion of space occupied by a 
physieal solid, or definitely determined in any way. Hence, 
also, a geometric solid im a limited y)ortiou of space. 

One advantage id UKing sEOmetrie solids liua in tlda. If we deidt 
with phyaioal Bolida oiUy, aa blocks oE wood, uuurMu, iron, etc., we 
should need to determine tlie jiroperties oJ each kind o£ physical aoiid, 



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]2 riiANE r.KOMF.TKV 

separately; but, by deterrainlug tiio jiropertits of a geomotric. nolitl, 
iwe deteimine onee for all the pvoperties of evoiy physical solid, tio 
matter what its material, that will exactly fill the space occupied by 
the giveu geometric solid. 

Hereafter in this book the term "solid" is understood to 
mean geometric solid, naless it be otherwise specified. 

7. Other geometric magnitudes defined as boundaries, 
A surface is the bonudiiry of a «oiid. 

A line is the boundary of a surfaiie. 
A point is the boundary of a line. 

The solid, surface, line, and point are the fiiijdr.mcntal 
geometric magnitudes, 

8. Geometric magnitudes defined by their dimensions. 

A solid has three dimensions; viz., length, breadth and 

A surface has two dimensions, length and breadth. 
A line has one dimension, length, 

A point has no dimension. Hence a point is that which 
has position, but no magnitude. 

9. Geometric magnitudes defined as generated by motion. 
Aline is that which is generated by the motion of a point . 
A surface is that which is generated by the motion of a 

line (not moving along itself). 

A solid is that which is generated by the motion at a 
surface (not moving along itself). 

The three independent motions by which a solid is generated ilhia- 



10. Geometric magnitudes as intersections. The inter- 
section of two surfaces is a line; of two lines is a paini. 

It is sometimes more advantageous to regard geometric 
magnitudes from one of the above points of view, some- 
times from another. 



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LINES 13 

11. A geometric figure is any combination of poiots, 
lines, surfaces, or solids. 

12. The form or shape o£ .i figure is determined by the 
relative position of its parts. 

13. Similar geometric figures are those whieh have the 
same shape. 

Equivalent figures are tliose which have the same size. 

Equal or congruent figures are those which have the 
same >jli<tpr and .■<izt\ iind can, therefore, be made to 
coincide. 

14. A point iri represented to tin- eye by a dot nnd ie; 
niimed by ll lett.et- Liftixed to the dot, as the point A, -A. 



15. A straight line is a line such that, if any two points 
in it be fixed and the line rotated, every point in the line 
will retain its original position. 

A straight line is also sometimes described as a line 
which has the same direction throughout its whole extent; 
or, as the shortest line connecting two points. 

The word "line" may be used for — - — ■- 

"straight line," if no ambiguity results. 

16. A curved line is a line no portion 
of which is straight. The word "curve " 
is often used for "curved line." 

17. A broken line is a line made up of different straight 

18. A rectilinear figure is'a figure composed only of 
straight lines; a curvilinear figure is a figure composed 
oTily of cnrved lines; a mixtilinear figure is a figure 
containing both straight and curved lines. 




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.14 



PLANE GI,OMKTr 



19. Kinds of straight lir.e. A .siraiKlit lin.- 
dcliTiite Of iudolinitc in Iciij- h. 

Tho line of (Jefinili^ k'li ;th is Hoiurtiincs b 
sef/inmi ot- m-cl. 

Otb^r kiiuta .>f RlraiKht line are .li.fiiicd in Arts. 2C an. 

20. Naming a straight line. A straight line i 

by naming two of its points, as the 

line AB (a sect) ; or the line CD (in- 

definite in length). A segment or sect 

may also be denoted by a single 

{small) letter, us the Hne a. J" ~ 



21. The circumference of a e 
of whieh id equally dirilaiit fro 
called the center. 



-de 



line every 
i point w 



ANGLES 



22. An angle i» the amount of opening between two 
sti'aight linos which meet at a point. 

The sides of an angle are the lines whose intersection 
forms tlio iin^jli'; the vertex of an angle is the point in 
which the sides intersect. 

23. Naming angles. (1) The most pr 
naming an angle is to use three let- 
ters, one for a point on each side of 
the angle with the letter at the vertex 
between these two, as the angle ABC. 

SiDce the aiae of an angle is indepen- 
dent of the length of its aides, the points 
nnmed on its aides may be taken at any 
plauB on its siilas. Thus, the angles AOl), 
BOD, iSO£,dOFa.ve all the same auglo. 




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ANGLKfi 

(2) In fiasc l.liei-c is Init onr iiiigk" at a eivoJi ven 
the lettor iit the vm-tox alone is sufri«ient to denote 
angle, as the angie in the last figure. 

(3) Sometimes a letter or figure phiced inside , 
the angle and near the vertex is a convenient ziL 
symbol, as the angle «. 



24. A straight angle is an !ii\<;li 
same straight Hue, hut which ex- 
tend in opposite direetions fi-oni 
the vertex, as the ang'le BOA. 



25. A right angle is one of two eniiiil 

one Ktr;tight liiiu hKiuliiig' ;uioll 
line I'Q meetw lino AB so a.^ to uial'ie 
angle PQA equal to angle I'QIi, iVM-h of 
these angles is a right angle. A riy^lit 
angle is also liiilf of a straight angle. 



whose 


sides li.' in the 




^ 




" 


ef|niil .1 
-ill lint 


nffles made hy 
. Thus, if the 



26. A perpendicular is a line th 
makes a right angle with a given line; 

thus PQ in the last figure is perpendicular to liA. The 
foot of a perpendicular is the point in which the perpen- 
dicular meets the line to which it is drawn, thus Q is the 
foot of the perpendicular PQ. 

27. An acute angle is an angle 
less than a right aiiijle, as the 
angle ^icr. 



28. An obtuse angle is a 
angle greater than a right angl 
hut less than a straight angle, l 
angle A OH. 



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IG 



I'LANE GEOMETP.V 



29. A reflex angle is an angle 
greater than a straight angle, but. less 
tlian two straigiit angles, as angle ^ Oi^. 

Ill tliU book, angles larger tliiin a atraigUt 
angle are not considered unless special mention ia ni;t.i 

30. An oblique angle is an angle tliiit is 
nor a straight iiiigle. Hence, oblique angle is 
for acute, obtuse, and reflex angles. 

31. Adjacent angles are angles 
whii^li have a common veiiex and a 
common side between them, an anglt's 
AOBan<iBOa. 

32. Vertical angles are angles 
which have a common vertex and 
the sides of one angle the prolonga- 
tions of the sides of the other angle, 
as the angles AOG and BOD. 




53. Complementary angles are two 
gotlicr equal a right angle, an the angles 
AOP a\>d rOQ. 

Hence, the complement of an angle 
is the difference between that angle 
and one right angle. 

u~ 

54. Supplementary angles urn two 

angles whicii together equal two I'ight /P 

angles (or a straight angle), as the an- y' 

gles A OP and POS. / _^ 

Hence, the supplement of an angle '^ 

is the difference between that angle and two right 
angles. 



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ANai.ns 17 

85. Angles as formed by a rotating straight line. If 
the line OB start in the position OA and rotate to the 
position OjB, it is said to generate tlie 
LAOB. ? 

The size of an angle may, therefore, \ I _,-' 

be considered as the amoniit of rotation ^ _ '^.j, .-'^ ^ 
of a line about a point, from the origi- /^ 

nal position of the hue. p/ 

If the rotation ia eon tinned far 
enough, a right angle {LAQG) is formed; afterward an 
obtuse angle {lAOD); then a straight angle {LAOli), 
and a reflex angle ( lAOV), etc. 

An advantage of this method of forming or coneeiving 
angles ia that by continning the rotation of the moving 
line an angle of indefinite size may be formed. 

36. Units of angle. A right angle is a unit of angle 
useful for many purposes. Sometimes a smaller unit of 
angle is needed. 

A degree is one-ninetieth of a right angle; a minute is 
one-sixtieth of a degree; a second is one-sixtieth of a 
niinnte. 

BesidoB tbuse, oDjuv units of angle are used for ceitain puipoaes. 

SURFACES. DIVISIONS OF GEOMEXEY. 
PARALLEL IINES 

37. A plane is a surface such that, if any two points in 
the surface be joined by a straight line, the straight line 
lies wholly in the surface. 

Hence, it plane figure is a figure sueh that all its points 
lie in the same j)laiif. 

38. A curved surface is a surface no part of which is 



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18 I'L.VXK f3E0M]':'Jliy 

39. Plane Geometry is thnt. bram^li of Geometry wljieli 
treats of plane figuves. 

40. Solid Geometry is that branoli of Geometry wLiiih 
treats of figures all points of whieh are not in tiie samei 
plane. 

41. Parallel lines are straight lines in tde s;ui;o piitno 
Tvliicli do not ]neot, liowever fur they be protUiood. 



Ex. 2 Which of the capital letters o£ the alphabet are str.iight, 
which curved, which broken, which curved aud straight linta 
coBihiiied t 

Kl. 3. Draw an acute angle. An obtuse angle. A refliix angle. 

Ex. 4. Draw two adjacent angles. ' Two vertical angles. 

Ex.6. What is the complement of an angle of 43"! What is its 
Bupplemtnt f 

Ex. 6, What is the complement of 57° 19' 1 of 62° 23' 43" 1 What 
is the supplement of each of these ! 

Ex. 7. At two o'eiocli, what k the angle made by the hour an.l 
minute hands of a clock 1 at three o'clonk f at five o'.'lo^-k ? 

Ex. 8. At 1 ;30 o'clock, what anf;lo do the hands of a clock raak,' ? 
at L>:I5t at 8;-lj? 

Ex. 9, What kind of an angle is the supplement of . an obtuse 
angle; of an acute angle f of aright angle t 

Ex. 10. What kind of a surface is the floor of a roomT the surface 
of a baseball? the surface of an eggt tlie surface of a hemisphere! 

Ex.11. If the surface JBCD n 
the right, what solid is generated h 



itt What surfaces are generated 
bounding lines T What 
tlcos of its angles t 



;nerated by ^~~S\ ' 

ated by its 1 

by the vet- Jr=^-:."v.::: 



■II 



---=.>a 



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GKOMETKIC MAGNITUDES ISI 

Ex. 12. Draw two supple me ntavy adjacent iicgles. Also two sup- 
plementary angles that are not adjaeent. Alao two adjacent angles 
that are not supplementary. 

Ex. 13. The sum of a right angle and an acute angle iswliat kind 
of an angle? Their difference is what kind of an angle! 

Ex. 14. The sum of an obtuse angle and a right angle is what 
kind of an anglel Their difference is what kind? 

Ex. 15. If three straight lines meet (but do not intersect) at a 
point in a plane, how many angles have this point as their common 
vertes? Draw a figure illustrating this and name the angles, 

Ex. 16. How many angles are formed if four lines meet ibut do 
not intersect) at a point in a plane! 



Es. 17. How many degree 


■s art' ii 


lan auglo whiph equals tw 


oomj.lement! 






Ex. 18. How many degret 


^s in ai 


a angle which equals ouf 


its supplement! 






Ex. 19. What kind of an 


angle i 


s greater than its suppk 



PRIMARY RELATIONS OF GEOMETRIC MAGNITUDES 

42. C<!vtjuii primary relations of geometric objects have 
already been giveu in the defiiiitioiis uaod for geometric; 
objecte. We now proceed to investigate the relations of 
geometrie maenifcudes more generally and systematically. 

43. An axiom is a troth accepted as requiring no 
demonatratioti. 

44. Two kinds of axioms are used in geometry: 

1. General axioms, or axioms which apply to o.'-.her 
Ivinds of rinaiitity as wi>ll as to geometric magiiitudesj for 
instance, to numbers, forcca, masses, etc. 



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liU PLANK GEOIIETHY 

2, Geometric axioms, or iisioms \ylii(!h apply to gee- 
metrii; tiiaguitiidtis alone. 

45. The geEcral axioms may be stut.t'd as follows; 

GENERAL AXIOMS 

1. Things H-hieh arc rqiud to fJir siime thintf, or to rqual 
ihingf, en' rqiiuJ fo Hwli nthrf. 

2. Jf rqnah be ciddfid to pcinalu, ihe niim^ are equal. 

.■J. // equals he subtracted from equals, the remainders 
are equal. 

4. Doubles of equals are equal; or. in general, if equals 
he multiplied hy equals the pndue/s tire eiii'ul. 

r>. Halves of equals are equal; or, in general, if eqnah 
he divided by equals the quotients are equal. 

6. The whole is equal to the sum of its parts. 

7. The whole is greater than any of its parts. 

8. A quaiility may he substituted for its equal in any 
proeess. 

9. Jfeqiials be added to, or suhtr acted from, miequals, the 
results are unequal in the same order; if uiiequals be added 
to Uiiequals in the same order, the results are unequal in 
that order. 

10. Doubles, or halves, of uuequals are unequal in the 
same order, 

H. If miequals he subtracted from equals, the remainders 
are unequal in the reverse order. 

12. If, of three quantities, the first is greater than the 
second, and the second is greater tJtan the third, then the 
jirsi is greater than the third. 



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GEOMETKIO AXIOHR 21 

46. The value of the general axioms. Tlie axioms given 
above seem so obvious that the student at first is not likely 
to realize their value. This value may be illustrated aa 
follows: 

II the distnnee fvom Washington to PhiladelphiB be known, and 
also the distance fvom Philadelphia to New York, the distance from 
Washington to New York may be obtained by adding together the two 
distaneea named; for, by axiom 6, the whole is equal to the sum of 
its parts. Thus the labor of actually meusuriiig the distauee from 
■Washington to New York is savod. 

Agaiu, if the lu'ifrht ot a suhoolboy o£ a ^'ivon age in Paris be 
measured, and the height of a like schoolboy in New York be meas- 
ured, and the result o£ the measnrempiit in each case is the same, we 
know that tho boys are o£ the same height, without the labor and cost 
of briugiugtlieboystogether and comparing their heights directly ; for, 
by axiom 1, things which are equal to the same thing are equal to 
each other. 

Thus the general axioms are to be looked at not merely 
as fundamental equivalences, bitt also as fundamental 
economies. For many purposes the latter point of view is 
more important than the former. 

47. Tlie geometric axioms may be stated as follows: 



GEOMETRIC AXIOMS 

1. Through Uvo given poinh onhj one slraighf line can 
be passed. 

2. A geometric figure may be freclij moved in space with- 
out any change inform or she. 

This aiiom is equivalent to regarding space as mu/orm, or homn- 
loidal; that is, as having the same properties in all its parts. It has 
already been assumed in some of the deSnitions given. See Arts. 15 
and 3u. 

3. Through a given point one straight line a'nd only one 
can be drawn parallel to another given straight line. 

By Art. IS, geometric figures which coincide are equal. 



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22 PLANE GEOMETCr 

48. Utility or uses oi the geometric axioms. By means 
of the firet geonintric axiom we urc able to ishrink or con- 
dense any striiight line into two points. Later tiiis atlvan- 
tage gives rise Lo many other advantages. By tlie second 
geometrie axiom, tbe knowk'djfe whicli ive liaye of ono geo- 
metric object may be transferred to anotlier like object, how- 
ever widely separated in space. The utility of the third 
geometric axiom can be miwle more evident wlieu we come to 
use it in proving ]iew geometriu truths. 

49. A postulate in geometry is a oonstruction of a geo- 
metric figure admitted as possible. 

50. The postulates of geometry may be .-stated as follows : 

1. Through any two points a straight line way he dt-airn. 

2. A straight Unetnay ie extended indejiHitelij , or it maij 
be limited at any point. 

3. A circumference may be described abont any given 
point as eenier, and with any given rifdiii.'^. 

These postulates limit the pupil to tbe hsb of the slrnight-eciKed 
ruler and tlie compasses in eonatmoting Gbuibb in geometry. One of 
the objects of the study ot geometry is to disi'over what geometrin 
figures can be coustructed by a combination of tbe eleiiienfnry eon- 
atraetions allowed in the postulates ; that is, by the use of tho two 
simplest drawing instraments. 

51. Logical postulates. Besides the postulates which are 
used in the actual construction of figures, there are certain 
other postulates which are used only in the processes of 
reasoning. Thus, for purposes of reasoning, a given 
angle may be regarded as divided into any convenient num- 
ber of equal parts. Whether it is possible actually thus to 
divide this angle on paper by use of the ruler and com- 
passes, is another question. 



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DEMONSTRATION OF PROPEItTlER '26 

DEMOKSTKATION OF GEOMETRIC RELATIONS 

52. A geometric proof, or demonstration, is a course of 
reasoning by whidi a relation lietween geometric objects ia 
established. 

53. A geometric theorem is a statement of a truth con- 
cerning geometric objects which requires demonstration. 

Es. The sum of the angles of a triangle equals two ri);'it angles. 

54. A geometric problem is a statement of the construc- 
tion ot a geometric iigiire which is required to be made. 

Es, On a given line to eonstruet a triangle toutalning three equal 

65. A proposition is a general term for either a theorem 
or a problem. 

Thus, propositions are sulidiviileil into two classes: 
1, Theorenas; 2, Problems. 

56. Immediate inference is of two kinds: 

1. Changmg the point of view in a given statement. 
Thus the statement, "two straight lines drawn through 
two given points must coincide," may be changed to "two 
straight lines cannot inclose a space." 

2. Reasoning which involves but a single step. 

Ex. "All straight angles are equal;" 

.'. "All right angles are equal." (Ax. 5.) 

57. A corollary is a truth obtained by immediate infer- 
ence from another truth just stated or proved. 

58. A scholium is a remark made upon some particular 
feature of a proposition, or upon two or more propositiona 
which are compared. 



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2-i PLANE GEOMETRY 

59. Hypothesis and conclusion. A |iroiiosition consists 
of two parts : 

1. The hypothesis, oi- Unit wlikh is known or granted. 

2. The conclusion, or that which is to he ]ii'ovi'd oi- coii- 
stvueted. 

Thus, in the pfoposition, "if two straight liiicH are per- 
pendienlar to the same line, they are parallel," the 
hypothesis is, that two given lines are perpendienlai" to 
jinother given line; the eoncUision is, that the two given 
I'mes are parallel. 

60. The converse of a pvoposition is another proposi- 
tion formed by interchanging the hypothesis and the 
conclusion of the original proposition. 

Thus, theorem, "every point in the perpendicular bisector of ii. ]itie 
18 equidistant from the estremitiea of the line;" 

Cotirefse, "every point equidiatant from the extremities of aline 
lies in the perpendicular biaeetor of the Hue." 

Or, in geneml, theorem, "if A is B, then A' is F." 
en»Terse, " if X is F, then A is nr' 

Frequently a converse is formed by interthanfiing p^i-t only of an 
hypothesis with part or all of the eoneluuion, or vice vi^rsii. 

Thus, theorem, "if ^ is B and C ia D, then J is I'r' 
com-erse, "" if J is B and X is Y, then C ia Jh" 

The converse of a theorem is not necessarily trne. Tims, 
it is true that all right angles are equal, but it is not true 
that all equal angles are right angles. 

61. The opposite of a theorem is a theorem formed by 
making both the hypothesis and the conclusion of the 
original theorem negative, 

Ex. theorem, "if A is B, then X is Y;" 

opposiff, "if A is not B, then X is not 1'." ' 



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DEMONSTRATION OF PROPEETIES 25 

If a direct theorem and its opposite ace both true, the uonverge is 
known to bo true without proof. 

Also, if a theorem and its converse are both true, the opposite ia 
known to be true withont proof. 

Thus certain economlea arise in the demonstration of theorems. 



62. Methods of geometric proof. Sevuciil principal 
methods of proving theorems are used in geometry. 

1. Direct demonstration. 

2. Proof by superposition, in which two iijjureiS are proved 
equal by makiiii^ one of Ihein eoiiieide wifh the other. 

3. Indirect demonstration, wliieh eonsists essfniially in 
showing that a given statement is true by i^howiiig that its 
negative cannot be true. 

Other special methods of proof will be pointed ont as 
they occur in the coarse of the work. 

63. FornI of a proof. The statement of a theorem and 
its proof eonsist of certain distinct parts which it is impor- 
tant to keep clearly in mind. These parts are; 

1. The general enunciation, which is the statement of 
the theorem in general terms. 

2. The particular enunciation, or statement of the 
theorem as applied to a particular figure used to aid the 
mind in carrying forward the proof. 

3. The construction of supplementary parts of the figure 
(not necessary in all proofs). 

4. The proof. This must include a reason for every 
statement. 

5. The conclusion. 

The letters Q. E. D., standing for "quod erat demon- 
strandum," and meaning "which was to be proved," are 
usually annexed at the end of a completed demonstration. 



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PLANE GEOMETi;V 



EXERCISES. CtiOi 



Es. 1, InthofiKurempas- 

ure AB: then meivsure BC. A — _— -a 

Now find AC without meas- 
uring it. Wliat axiom hiive you used 1 

Ex. 2. If /L ^0^ = 00°, I BOC=m'' and ■'; /''' 

ZCOi) = 130°; find without measui-ing them a ~X / 

AOC and BOD (reHe.";). What axiom have you N, / 

Ex. 3. Prove by repeated use of the first 
part of Asiom 1 that magnitudes equal to equal [ 

magnitudes are equal to each other; (thus, I> 

given J=!.r, li=y, and x~'j. Prove A^B). 

Ex. 4. Give a namevipal iliuKtration of Axinm 11, 
Ex. 5. Show liy ttia axioms that a part is equal to a whnlo di- 
minished by the remaining part, 

Ex. 6. Show that Axiom 1 is a special chsb of Axiom K, 

Ex. 7. It a = x + }j and ;c = y, show by use of the axioms that 

Ex. 8. Di'aw a line and produce it so that the produced part shall 
equal another given line. 

Ex. 9. By use of the eompasaes, mark off on a given line a part 
twieo as long as another given line. 

Ex. 10. On a givenlinemarkoff.byfoive^t uses of theoompHSses, 
a part four times as long as another given liue. 

Ex. 11. Di^aw three straight lines and denote them hy I, ra, 
andii. Then draw a line l + ni-~n, and also a line I — 2'i(+3ii. 

PROPERTIES OF LINES INFERRED IMMEDIATELY 



64. If two straight lines have two points 
the lines coitidde throughout their whole extent (Art, 47, 
Geom, As. 1). 

Hence, two straight tines can intersect in hut one point. 

65. If two straight lines coincide in part, they coincide 
throughout, 



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PEOPERTIES OP ANGLES 27 

66. Only one straight line can be drawn eonnecHng two 
given points. 

67. Two straight lines cannot enclose a surface. 

68. A given straight line {sect) can he divided into two 
equal parts at but one point. 

For (by Ax. 5) halves of equals (or of the same thing) 
are equiil. 

PROPERTIES OF ANGLES INFEEEED IMMEDIATELY 

69. All glrnight angles arr equal. 

70. A straight angle can he dividfd into lieo equal 
angles by but one line at a given point . 

in the given straight Zme. 

For (Ax. 5) halves of the same mag- 
nitude are equal. 

71. Hence, at a given point in a straight line hut one 
perpendicular can he erected to the line. 

72. All right angles are equal. 

For all straight angles are equal (Art. 69) and halves 
of equals are equal (Ax. 5). 

73. The sum of the two adjacent angles / 

formed by one straight line meeting an- F^i 

other straight line equals ttvo right angles. 

For the angles formed are supplementary adjacent 
angles (Arts. 31, 34). 

74. If two adjacent angles are together equal to a 
straight angle {or two right angles), their exterior .Hid«'ts 
form one and the same straight line. 

For their esterior sides form a straight angle, and 
hem;e must lie in a straight line (Art. 24). 



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28 



}'T.ANE 03?,OJIETRY 



75. The cowplemeiiix of Iwo equal nn^jlr-: /nr equal 
(Art, 72 and Ax. 3); the aiipplemetits 
of two equal angles are equal (Art. Gi), 
Ax. 3). 

76. The sum of all Ike. oiujles about a 
point equals four rhjhl amjleti. 

Thus, la + lh + lc + ld^-le--^ 
4 rt. A.. 

■77. The SUM of all the angles about 
a point on the same side of a straight 
line passing through the point equals 
two right angles. 

Thus, Zj3 + Z5 + Zr = 2 i-t. i. 




EXERCISES. CROUP 3 

Ei. 1. How many different straight linps are determineii liy tliree 
points not in the Same straight line? 

Ex. 2, How many strniifht lines are determined by fouc points in 
a plane, no three of them being in the same stra:(;(it line ? 

Ex. 3. If, iii Fig. 2 above, Aa,b,c. cl^ 40°, 50°, 60°, 70° i-esppp- 
tively, find Z e. 

Ex. 4. IE,- in Fig. 3 above, the lines forming the an^ld q are per- 
pendieulat to each other and I p = 47°, find the other angles of Iho 
fignre. 

Ex. 5. Measure Zrt o£ Fig. I on preceding page. Find Zb with- 
out measuring it. Now measure Ih and 
compare the two results, , A 

Ex. 6. Given QB1AH, PB 1 BC, and \ ,q 
IABC = 1W; find tha otliot angles of the \ ^^ 
figure. \l rrr\' 

B a 

Ex. 7. Arrinpe file points in ^ pKne 
so that the few ! st n imber of st light line'! i av pass through them, 
no line to pass through more than IhrfiL [ lul? 



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PLANE GEOMETRY 

Book I 
RECTILINEAR FIGURES 



Proposition I. Theoreji 
78. If one simight line inUrsecfa anotlipr siraight line, 
the opposite or vertical angles are equal. 




GiveE the straight linos AB and CD iutersoetiiig at 
the point 0. 

To prove ^AOC= IDOB and IA0T)= ICOli. 

Proof. /. AOG+ ZA0D=2 rt. A, Art. 73. 

[tkv Kiim (jf (ICO ndjaecut angles formed hij one striikjht hue meclimj 
aiiullicr straight line equals tu-o riylit angles). 

Also ZfiOD+ZJ.OD==2rt. A, 
isame reason). 

.-. ZAOC+^AOD=^lBOI>+ZAOD, A^i. i. 

{things equal lo the same ilibuj are eqiiul to t'udt other). 

Subtracting ZAOD from the two equals, 

/.AOG = lBOD, Ax. 3. 

[ifeijiials he siiblraotcd from equals, the remainders are equal). 

Id liliu iiiamu;r it may be proved that 





IA0J) = IC0B. 


O.B. B. 


Ex. If i 
without uihu 


n the above 6gure IDOB^IQ". 
isuriug them. 


Cud the other angles 



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30 



BOOK I. PLAXE GEOMETltV 



Proposition; II. Tiieoeem 
79. If, from a point in a perpendicular' to a given line, 
two oblique lines be drawn cutting off on the giceu line equ<tl 
sffimenia from the foot of the perpendicular, the oblique lines 
are equal and make equal aiujle.s with the perpendicular. 




Given a line AB witli CI) ± to it at the point C. and 
FR and FQ drawn from :iny point as P in CD, cutting off 
CR = QCon AB. 

To prove FR ~ FQ and l CFR = l CFQ. 

Proof. Fold over the iigiive DCB abont J}C as an axis 
till it i;omes in the plane DOA. Geom. Ax. 2. 

Then ^DCB = /. DCA {allrighl A are = ). Art. T2. 

.■. line CB will take the dii-eetion of CA. 



But 

Hencf 

And 



CJi= CQ. 

.: point R will fall on point Q. 

ine PR will coincide with line FQ. 



Hvp. 



■u-aiyia 






ectiu'j U 



I CFR will coincide with / CFQ. 
.: FR = FQ, and Z CPE = I CFQ, 
{geometric figures which coincide are equal). 



Ex. 1. Point out the hypothesis and the oonelusioii ii 
eounciatton of Prop, I. Also point thorn out in ti 
enunciation. Do tho same tor Prop, II. 

Ex. 2. If three straight lines intersect at a point, 1 
the angles formed is it necessary ti 
all the augleat 



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LINES AND ANGLES 31 

pROPOsiTioK III. Theorem 

80. From, a given point wUhftut a straight line but one 
perpendicular can be drawn to the line. 



Given the str,iif;iit lino AB, P any point witlioiit: AP,, 
PQ ± .1 /i, iiud PR any othet- line dniwn from /* to .4./-!. 
To prove that PR is not ± Ali . 

Proof. ProdneePQ to P' making Q/'' = i*Q. Ih-a.w RP'. 

Then RQ 1 PP'. Tiyp. 

P'Q^PQ. Constr. 

.-. ieP = RJ^ and Z PiiQ = Z P'ii^g, Art. 79. 

(*/. /'"o»i a ;w/h( t» a ± (0 a ghea line, tvto ohligne lines be drawn 

cattiiiij o^ on the gii\n line eqiiiil segments from XMftiot of the X,, 

ike oliliqae lines are eqaal anil make equal A with the L ) . 

Bnt Pif^F is not a straight line, Art. G6. 

(only one slniiijht line can be dnuvH eiiiiuecUng tiro gii'en piiiiiln), 

:. PRI" is not a stmiglit Z . 

.-. Z PRQ, the half of Z PRP, is not a rijjlit /. . Ax. 10. 

.-. PR is not ± AP. 

.'. onlv one perpendicular can be drawn from P to AP, 

Q. E. ». 



Ex. Tlll'BB 


traiRlit liii 


a iiiterse 


t tit a pi) 


aiiglos lormod 


It tlie iKtiii 


Me 30° 


XQd 4U". 


ftt the point. 









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I'LANE (iKUMF/l'RV 



TRIANGLES 



81. A triangle is a portion of a plane homideil 1 
straight liiiow, as the ti-iaiigle AUG. 



82. The sidM of a triangle are 
the lines which bound it; tlie 
perimeter of a triangle is the sum 
of the sides; the angles of a tri- 
angle are the angles formed liy 
the sides, as the angles ^1, B 
and G; the vertices of a triaui>:le 
angles of the triangle. 




83. An exterior angle of a triaiigb 
one side and by another side 
produced, as the angle BCD. 
With reference to the 
BCD, the angles A and B 
are termed the opposite inte- 
rior angles. 




84. Classiiication of triangles according to relative length 
of the aides, A scalene triangle is a triangle in whieli no 
two sides are equal. An isosceles triangle is one in wlueh 
two sides are equal. An equilateral triangle is one in which 
alt three sides are equal. 




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TEIANGLES 



33 



85. Classification of triangles with reference to character 
of their angles, A right triangle is a triangle one of whose 
angles is a right angle. An obtuse, triangle is a triangle 
(me of whose angles is an obtuse angle. An acute triangle 
is a triangle all of whose angles are acnte angles. An equi- 
angular triangle is one in which all the angles are equal. 



86. The base of a triangle is the side upon which the 
triangle is supposed to stand, as AB. The angle opposite 
the base is called the vertex angle, its 
angle ACB\ the vertex of a, triangle is 
the vertex of tlie vertex angle of the tri- 
angle. 

The altitude of a triangle is die 
perpend iciil.'ir from the vortex to the ^ 
base or base extended, as CD. '-' 

87. In an isosceles triangle, the legs are the equal 
siilys, and the base is the remaining side. 

88. In a right triangle, the hypotenuse is the side oppo- 
site the right angle, and the legs are the sides adjaeent to 
the right angle. 

89. Altitudes, bisectors, medians. In any triangle, any 
side may be taken as the base; hence the altitudes of a 
triangle are the three perpendiculars drawn cue from each 
vertex to the side opposite. 



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-1-i BOOK I. I'LANR fiEOMl?T!!Y 

A bisector of an angle of a triaiiglo is a line wliicli ilividcv 
tliis angle into two equal pavte. ThJH Liswtur is usiiiLlly pru- 
diieed to mout tlto side opposite tlio givwi aiiglu. 

A median of a triangle is a line drawn from a vertex of 
the triangle to the middle point of the opposite side. How 
many medians has a triangle ? 

90. Two mutually equiangular triangles are triringles 
having their corresponding angles equal. 

91. Homologous angles of two mntnally equiangular 
triangles are corresponding angles in thnwe triangles. 

Homologous sides of two mutually equiangular triangJes 
are sides opposite homologous angles in those triangles. 



We shall now proceed to determine first, the properties 
of a single triangle, as far as possible, then those of two 
triangles. 

92. Property of a triangle immediately inferred. The 
SUM of any two sides of a triangle is giratfir than ihf third 
mJp. For a straight line is the siiortest line between two 
points (Art. 15.) 

Ex. 1, Point out the hvpotiiesis and eonelusion in the generd onun- 
tiation of Prop, 111 ; also point tLtm out in the pnrtienirLi' enunciation. 

(As each of the uoxt Cftuen Props, is studivJ, let the pupil do tlia 
same for it.) 

Ex. 2. Find tlie angle whose complement is IS"; whose supple- 

Ei. 3. It the eomplement of an angle is known, what is the  
shortest way of finding the supplement of the auglef If the supple- 
ment is known, what is the shortest way of finding the complement ? 

Ex. 4. In 25 minutes, how many degrees does the minute-hand 
of a clock travel 1 'Rtm m.iuy docs the honr-hand f 

Ex. 5. Draw three straight lines ao that they shall i 
three points; in two points; in one point. 



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TEUNGLES SJ 

Proposition IV. Theorem 

93. Aiiif aide of a triangle is greater than the difference 
between the other two sides. 



Given A B any side of the A -4 liC, iiiul A € > iiC. 

To prove AB>Aa—nC. 

Proof. AB + BOAC, Art, 93. 

(the sum ofawj two sitlcs of a triangle ?« i/rcaler lha» the thinl side). 
Subtracting BG from eacli member of tlie itiequalitj-, 

AB>A€—BC, A^.!). 

'Sf equals he subtracted from unequals, the remainilers are uneijual in the 
same order). q, s, D. 

94. Cor. TJie perpendicular is the shortest line thai 
can be drawn from a given point to a given line. 
For, in the Fig. pnge 31, 

PP'<FB + RI", Art92. 

Or, 2 1'Q<2rR. Ax. 8. 

.-. FQ<PR. A>^.V). 

Hence, Dep. Tlie distance from a point to a line h the 
perpendicular drawn from tjie point to tlio line. 

Ex. J, If one sido of an equilateral triniifjle U 4 iiioliea, n-iiat i^ ita 
perimeter ? 

Es. 2, la it possible io form a triangle whose sides are 6, £1 and 
17 inehea f Tt7 to do this witk the compaaaea and ruler. 

Ex. 3. Is it possible to form a triangle in whii/ii one side is 10 
inches and tho diiferein'o of tLe other two sides is I'l Inches f 

Ex. 4. On a given line ns base, liy e.xm^t uae of ruler and uoni- 
pftsaos, eonstrucL au equilateral tviauglo. 



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VLAXE GEOAn'TltV 



PltDI'OSITlON V. TlIKOliKM 



95. If, from a point inth'm a irimu/U, i '■'■■ an ' ''■ nri'm 
fo the extrfmilies of otie side of the triangle, Ike ^'ihi of 
ihe other tiro xides of the trimigle is greater than the sum 
of the two Uii';:^ no drawn. 




Given P nny 
FA and I'G lines 
side AG. 



lit within tho tvi!\iislfi ABO, aw\ 
WW from r to the extremities of the 



To prove AB + BC> AF + PC. 

Proof. ProaiicR the line AF to meet EC at Q. 

Tlien AB + BQ> AP+ FQ. A 

(a siraujIU hue !.s tlie siiortei line coiuieclMg tuo poiul^) 

I Also FQ + QG> PC, 



(s,i 



n>). 



Adding tliewe inecinnHties, 

AB + BQ + PQ + QC > AP+ FQ + PC. 
Substituting BC for its equal BQ + QG, 

AB+ BC+PQ> AP+ FQ + PG. 
Bubtraetiiig PQ from each side, 

AB + BC> AP + PC. 

Ex. Ozi a given line as base, by exact use of ruler and c 
onBtvutit au ia03cel«H triangle each of whose lega ia double 



. E. D. 
irapassea, 



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TKIANGLES di 

Proposition VI. Theorem 

96. Two triangles are equal if two sides and the in- 
eluded angle of one are equal, respecthebj, to two sides and 
the included amjle of the other. 




id />/;/'' in which AB = I)E, 



Given tlie triangles ^l HU a 
AC^BF, and lA^ IP. 

To prove A ABC = A DEF. 

Proof. Place the A AB(7iipon the A DEF so that the 
liue AG iioncide.s with its equal DF. Geoui. Ax. L'. 

Then the line AB will take the direction of DE, 
if,.rlA^ZDh!i h,j,>.). 

Also the point B will fall on E, 

(for line AE = line DE hj hyp.). 

Hencfi the line BO will coincide with the line EF, Art. 6G. 
{i^tlij one stiaiglil line oaH he drawn coimecting two yuiiU:,). 

:. A ABC and DEF coincide. 

.-. A ABC = A DEF, Art. 47. 

(geometric figures lirRicft coincide are equal) . 

q. E. D. 

El. 1. What kind of pvoof is used in Prop. VI t (See Art. 02). 

Ex. 2. IE S A, Baud C=CO°, 70°, 50°, J/J=10, Ji7=19, BC=iS: 
also Z 7>=00°, I}E=\6, DF^IO: find -i E and F and side EF 
wit bout measuring tbem. 



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liOOa I. I'l.A 



VnoroHvnas VII. TiiEOiiKM 

97. Ttco ifUmi/l-'s are e-iiml if two anijli'^ 'ind fhr h,. 
eluded sid<! of one ay cquiil, respet-ticdy, to iim angles and 
the included side of the other. 




Given Llic A ABC iiiid DEF in which ^A=ZD, 
IC= IF, i\uAAG=llF. 

To prove AAnC=ADEF. 

Proof. Plaee the A ABG xipon the A I>EF so tliiit AC 
shall coincide with its equal DF. Gsom. As. 2. 

Theu AB will take the direction of DF. 

(for lA^lDhyluji,.), 

and the point B will fall somewhere on the line 1>E or DF 
produced. 

Also the line CB will tal^e the direction of FF, 
{for LC = lFhjliyp.), 

and the point B will fall on FE or FF prodnced. 

.". point B falls on point F, Art. f>4. 

{two straiglU hues ciin intersect in but one jioiixi]. 

.'. &. ABG and DEF coincide. 

.-. AABC^ADEF, Art, 47. 

{geometric figures toliich coincide are equoX). n e, B 

Ex. 1. What kind o( proof is used in Prop. VII. ! 

Ex.2. If A A, B, 0=65°, 55°, GO', AB=2i. Aa=-['i, BC-2T: 
alio 4 D, f=C5°, 60°, and J)f=18; find DE, EF, and I E, 

Ex. 3. Conatriict by exiat iisa of ruler and compasses a scalene 
triaujjle whose sides are 2, 3 and 4 times a givun line. 



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TMANGLES Cif 

Proposition VIII. Theorem 

98. Two right iriangUs are equal if the hijpotenuse and 
an acute angle of one are equal to the hypotenuse and ait, 
acute angle of the other. 




Given the riglit & A BC aiul DEI'^ in wliieii liypote- 
nuse .4.iJ = liypotoimse DE, iiud lA = Z 1). 



To prove 



AABG=£X TiEF. 



Proof. Pkfie tlie A ABG npon tiie A I>EF so tliat the 
side AB shall coincide with its equal, the side DE, the 
point A coinciding with the point D. 

Then the line AC will take the diret^tion of DF, 
{for lA^ZDbuhnp.) 

Also the side BO will coincide with the side EF, Art. RO. 

{from a given point, E, viithont a straight Itne, DF, but one ± ean be 

(Irawn to the line). 

:. A ABC and DBF coincide. 

:.AABC=A.DEF, Art. 47. 

{geometric figures ■which cuinckie are eqnal). 



n equilateral trmngle, n 



\. E, D. 
li of whose sidea 



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HOOK I. I'LANE GKOltETRY 



Proposition IX. Theorem 
99. Ill on isosceles trinn'jle the angles opposite- the equal 
sides are equal. 




Given the isosceles A ABC in which AB=^BC. 

To prove ZA = /.C. 

Proof. Let BD be drawn so as to bisect /.ABC. 

Then, in the & ABD and DBG, 

AB^BC. Hyp. 

Also BD=BI), Ide»t. 

And IABD= IGBD. Constr. 

.-. A ABD- A CBB, Art, 96. 

wo & are eqaiil if two sides and the included Z of one are egual, respee- 
Uwly, to two sides and the included Z of the other. ] 

.: /LA=lC, 
{homologous A of equal A). 



Ex. 1. On a given line as base, conBtruot exactly an equilateral 
triangle above the line and another below it. 

Ex, 2. On a given line as base, eonstruet esactly an isosceles 
triangle whose leg shall be equal to a given line; make tlie same con- 
Btrnetion below the given line and join the vertices o£ the two 
lEOaceles triangles. 



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TEIANCLER 41 

Proposition X. Theorem (Converse of Prop. IX) 

100. // two ait(]leK of a triangle arc equal, Iha xidei 
opposite are equal, and the triangle is isosceles. 




Given Ap frlanfflc ABC in which ^A=-- Z BCA 
To prove AB = BC. 

Proof. U thu sides AB and BC are not equal, 
them must be longer than the otlier. Let AB be 
thac BC. 

On ^Sniark off ,10-BO Dmw DO. 
Then, in the A ABG and ATJC, 
AD^BC, 
AC=AG, 
IBAG^ IBCA. 

((wo Si. are equal if iico siijes and the included Z of: 

reapcctircly, to tico sides and the included L of the otlier) . 

Ora part is equal to the whole, which is impossible. As. 7, 
Hence AB cannot be greater than BC. 
In like manner it may be shown that AB is not less 
than BC. 

Hence AB=BC. 



Art. 90. 



What ineti.od of lirooE 
It ill a tri!iiigle VEF, 
Dtaw u figure nui oil it 1 



i usi-d hi Prop. X f 
ai'fc the value o£ tbe I'ai 



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BOOK I. PLASE GEOHlETIiV 



Proposition XL Theorem 



101. Tim triangles are cqmil if the thrrr .w'.'.s- "f mis 
•e equal, re^pi'i'lirchj, to the- three sides nf tin- <)(h:r. 



Given the A ABC iiiul BEF in wliieli AIi = l)E, I1C= 
J3F, fnidAG^nF. 

To prove A /U.'C= A BEF. 

Proof. Place the A ABC so that its longest siile AC 

shall eoineidu with its equal DF in the A DEF and the 

vertex B shall fall ou the opposite side of DF from E. 

Geora. As. 2. 

Draw the line EB. 

Then BE^VB (Hy-p.) .: A DEB is isosceles. Def. 

.-. Ip^ /: r. Art. !)9. 

(in an isosceles A Wie A opposite the equal sidun are equal). 
In like manner, in the A BEF, tq=- Ls. 
Adding, 

L'P^ Lq= Zr+ Is, Ax. a. 

Or II)EF=^ Z DBF. Ax. ii. 

.-. A DEF = A DBF, Art. 9G. 

{tKo A arc equal if Iwo Hides ami the iHi-hutei Z of one are c<iiial, 

respcctirclij, to Iwo sides ami tlie inrbuled I of Die other). 

:. A ABC = A BEF. Ax. 1. 

  — . —     Q. E. D, 

Ex. 1. Construct two equilateral triangles on the 
same base, one abore and the other helow, and join the 
two vertices. Prove that the line joining the vertices 
bisects t1ifi Tf-rie^ iingl"a, and also bisects the base at 
right angles. 

Ex. 2. Hence, at any point in a given straight line, 
construct exactly hy use of ruler and oompasseB a 
perpendicular to tliat line. 



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TRIANGLES 43 

Proposition XII. Theorem 

102. Two right triangles are equal if the hypotrniise and 
a leg of one are equal to the hypotenuse and a leg of the other. 




Given two right A ABC and Dl^F l!avm<; the hypote- 
nuse AB =h3-poteimse DE, and BC=J-iF. 
To prove A AB€= A BEF. 

Proof. Place the A ABC m that BG shall coincide -with 
its equal, EF, and A fall on the opposite side of EE from 
D, at A' . Geom. As, 2. 

Then A'F anA. ED will form a straight line, A'FI>, Art. 74. 
(i/ tmo adj. A are toqetticr equal to two rl. A , their ej:t. sides form one 
and the same straight line). 

A'E^EIK Hyp. 

,■, AA'ED is isosceles. Dt.>f. 

;. /1A'= Z.D, Art. 99. 

^oscelcs A the A opposite the equal sides are efiiiaL) 

:. A A'EF= A DEE, Art. 98. 

are eqiialif the hypotenuse and an amie i. of oi 
the hypoteniwe and an acute I of the other). 

AABG=ADEF. a^- 

__-__^_^~_ '■ ^- "• 



But 



(i« 






ray given stcaigLt 



Ex. 2. Biaect any yivBii imgk AOB, 



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BOOK I. PLANE GEOMETRY 



rROl'OKITlOX XIII. ThEOEEM 

103. All erierior iniijJe of a iriangk is greater than 
thcr iippiixite interior anf/k. 



Given / BGI) an esterior Z. of the A ABC. 
To prove Z SOD greater than lAJWor /BAG. 
Proof. Let E be the middle point of the line IIC, 
Briiw AE and produce it to F, making FE^AE. Draw 
FC. 

Then, in tlie A AEB and EEC, 

AE^FE, and BE - CE, Constr. 

I BE A= Z. FE€{Mng vertical i). Art. 78. 

.-. A ^ J:B - A FEC, . Art. %. 

(frco & are (qual if lico aiih-s and the indiiilril I of one are egtial, re- 

spet-lk-ciy, to two iiik-s and the indmlcd I of the other). 

:. ZABE= IFCE, 

{being bomologmis d of eiiiial A]. 
But Z BCD is greater than Z FCE, Ax. 7. 

{the ic/iD?e is greater than any of its parts). 
Substituting I ABE for its equal ZEOE, Ax. 8. 
ZBCB is greater than I ABE, that is, than lABG. 
Similarly, by drawing a line from B through the midpoint 
of jj (7 :ind by producing SC through (? to a point .ff, it may 
be shown that lACH {= IBCD) is greater than ABAC. 
Q- E. D. 

eacU of wLose legs eqaala balf a given line. 



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TRIANGLES 45 

PROPOsiTioN XIV. Theorem 

104. If two sides of a triangle are unequal, the angles 
opposite are unequal, and the grette-" angle is opposite the 
greater side 




Given the sida BC > side AB in the A ABO. 
To prove Z BA C greater than Z ('. 



Then, in the isosceles AABD, Z)'= Zs, Art. 99, 

{in an isosceles A the £s opposite the equal sides are equal). 

ZjB^(7 is greater than Zr, As. 7. 

[the Kliole is greater than anij of its parts). 
.■. ZJBAC is greater than Zs. Ax. K. 

But Z.s is ,111 exterior Z of the A ABC. 

:. Z/{ is Ki'eater than Z C, Ait. loa. 

(anext. Z of a A is greater llnm eUltor opposHe iiil. Z ]. 
Much more, then, is Z BA€ {which is greiiler than Zs) 
greater than Z C, Ax. 12. 

(if, iif thrre (fiinniities, the first is greater Ihnn the secmid, OHii the Keeoiiil 
is greater than the third, then the first is grenler than the Ihu-il) . 
Q. £. D. 

105. Note. The esaeutial afeps of the above proof may be 
arranged in a single statement, tUua: 

lJ<AlJ>Zr = ls>lV .: IBAC is greiiter than IC. 

Ex. J. Which is the longest side o£ a riglit triangle ? of an obtuse 
triangle f 

Ek. 2. Conslriiot exactly an equilateral triaugle, Bttth of ivhoee 
Bides is half a ^iveu line. 



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PLANE CEOME'l'HY 



PKOi'osiTiON XV. Theorem (Converse op Peop. XIV) 
106. If two angles of a triangle are unequal, tie sides 
opposite are luieqiial, and the greater side is opposite the 
greater angle. 



Given I A greiitei- th;iii Z C in the A ABC. 
To prove BC > AJi. 

Proof. BC either equals AB, nv is less than .4 B, or 
greater than AB. 

Biit BC cannot eqna! AB, 

for, if it did, /.A would equal Z C, Art,' 
(being opposite equal sides in an isoscvles A), 
But this is contrary to the hypothesis. 
Also BC cannot be leas than AB, 

for, if it were, Z A would be less than Z <7, Art. i 
(i/ two sides 0/ n A aye unequal, the i amiosite are uueqattl, ami 
greats Z ia opposite the tfreater side). 

This is also contrary to the hypothesis. 
;. BOAB, 

(fzr it ncilhcr equals All, uor in !ras than AH). 

Q. E. B. 

El. I. Draw a triangle the altitude of whieh fuUa on the b 
produced. What kiud of a triangle is tliis i 

Ex. 2. Draw a triangle the altitude of 
which coincides with one side. What kiud of 
a triangle is this f „^ — .^^ 

Ex. 3. By exact use of the ruler and com- 
passes, dww a perpendicular to 6 given liue 
from a given point without the line, ' " 



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TMANGLErt 



PftorosiTlON XVI. T1IEORH.M 



107. If two triangles have two sides of one equal, respec- 
tively, to two sides of the other, but the included angle, of 
the first greater than the included angle of the second, then 
the third side of the first is greater than the third side of 
the second. 





Given tli(! A AIW and J)EF in ivincli AI1 = DE, 
BC=EF, and lAtW is greater than I.E. 
To prove AC > liF. 

Proof, Place the A I)EF so that the side DB coincides 

with its equal, the side AB, and I*' takes the position F' . 

Geom, As. 2. 

Let the line BE bisect the IF'BG and meet the line 

AC at H. Draw F'H. 

Then, in the A F-BH and BBC, F'B^BC. Hyp. 

BH^ BE, Went. 

IF'BE^ ICBH. ConstT. 

.-. A F'BH'^ A BEC. (Wby?) 

.', F'E=CH, {homolot/ous sides 0/ equal ^). 

Hut AR + EF' > AF", Art. 92. 

{Die Slim oro'll ''f skies 0/ a A is greater than the third shU], 

SubslitutiiiK for EF' its equal EC, 

AE + EC, or AC> AF'. Ax. a. 

:. AC > 1)F. A.,. 8. 

q. E. ». 

Ex. 1. Draw a figure tor Prop, XVI in which the sides and au- 
gles are of sneh a size that i'" falls within tlie trianftle J fir. 
Ex. 2. Uraw another figure in whicli *" fiUU on the side AC. 



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43 llOOK I. PLANE GEOMETRY 

ritOfORITlON XVII. ThBORE.1[ (CONVERSE OF PUOI'. XVI) 

108. // tico sides of a irimigh are equal, respectivehj, 
to two aides of nnotJier trimigle, but the third side of the. 
fii-xt is greater than the third side of the second, then the 
unfile opposite the. third side of the first triangle is greater 
titan the angle opposite the third side of the second. 




Given thp A ABC iind HJ^F hiwing AB^-DE. B0= 
EF, but AC > DF. 

To prove ZJJ greater tliuii ^E. 

Proof. The Z.B either equals Z.E, or is less thunZE, 
or is greater than Z E. 

But Z B does not equal ZB, 

for, if it did, A ABC -would = A DEF, Art. or,. 
{tiro A are eqiial if tico "iilcs and the ineluileil I of one are equid, 

rcspeclh-etij, to tiro sides ami the iiieluihil Z of the other), 
and. AC would eqiiiil DF (Jtomoiogoua sides of equal &^) , 
which is contrary to the hypothesis. 

Also if AB were less than Z£, 

side AG would be less than side DF, Art. 107. 
(if two A liavetwosides of one equal, respeelivelji, io two sides of the olhtr, 
hut the itictaded Z of the first greater tliait the iacludeil L of 
the second, then the tAird side of the first is greater 
than the third side of the second). 
But this is also contrary to the hypothesis. 
Hence ZB is greater than ZE, 

{for it neither equals I E, nor is less than ZE). 



Q. E. 9. 



Ei, On a f-iven line (0 con- 
nictt a triangle whose other two siiles 
■e equal to two giyeu lines {m and »). 



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LINES 49 

PfiOPERTIES OF tlHES PROVED BY *ISE OF TSIAWGIES 

PRorosiTioN XVIII. Theorem 

109. Of lines drawn from the same point in a perpen- 
dicular and euttinij off unequal segments from the foot of 
the perpendicular, the more remote is the greater. 




Given PO 1 AB, FT and PQ oblique to AB, aud 
OT > OQ. 

To prove FT > PQ. 

Proof. J'roani^e PO to the point P' making OP'=OP. 

Oil AB take OR=OQ. Draw PE, P'Q. P'R, P'T. 

Then I'Q = PR, Art. 79. 

{if^froin a pnint in n ± , a gii-cit line, two oblique linen be drawn, mitliiig 

off oil the gieai line ri/iiat segmtnts from the foot of the X , the 
oblique lilies are equal). 

Ill A PTI", PT+ TP'>PB-VRP', Art. 93. 

[if, from a point u-illiin n A, two lines be drmnt to the extremilies of a 
side of the A, the sum of the other two sides of the A is greater 
than the sum of the tm lines so drawn). 

But oris X PP and PT Mid P'T cat oft equal segments, 
PO and i:"0, fi-crni the foot of the L AO. 

Hence PT^ P'T. In like" manner PR = P'R. Art, 7£». 
Hence, by substitution, 2 PT > 2 PR. Ax. 8. 

.-. PT > PR. Ax. 10. 

.-. PT > PQ. Ax. a. 



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PLANE GEOMETRY 



Proposition XIX. Theorem (Converse of Prop. II) 

110. Equal ohlique lines drawn from apoiiU in a per- 
pendicular cut off equal segments from the fool of ike 
perpendicular. 




Given PC ± AB, PK and FT oblique to AB, and 
PR = FT. 

To prove CE=Cr. 

Proof. Ill the right A RFC aiul CPT, 

PG^PG. Ident. 

Also PE^PT. Hy,. 

.-. ARPG^ACI'T, Art. ir- 

{tico right iJi are equal if ihehijpoleimse ami a leg of one are equal !•• ir,. 
hypotetiuae and a leg of the other). 

:. RG=CT, 
{liovtologotis H'ulei) of eipial ^). 

1). E. B. 

111. Cor. Of two unequal lines draion from a poini 
in a perpendicular, the greater line cuts off the greater 
segment from the foot of the perpendicular. 

Thus, if PT> PQ (Fig. of Prop. XVIII), OT cannot 
= OQ (Art. 110) ; nor is OT < OQ (Art. 109) .-. OT > OQ. 

Hence, also, from a given point only two equal straight 
lines can be drawn to a given line. 



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LINES 51 

Proposition XX. Theorem . 

112. I. Efery point in the perpendicular bisector of a 
line is equally distant from Ihe extremities of the line; and 

II. Eeery point not in the perpendicular bisector is 
unequally distant from the extremities of ihe Una, 




Given the middle point of the line AB, OC ± A P.. 
P any point in 00, and Q any point not in OC. 

To prove AP^^PB, bnt QA and QB unequal. 

Proof. I, AP^PB, Art. 79. 

(if, from a point in n A- to a gii:en line, tifo ohliqiie lines he drtiwit 
cutting off OK the given line equal cutji'iinttn, eb:.). 

II. Since Q is not in the line OG, either AQ or QB must 
cut the line OG. 

Let AQ intereeet OG in the point R and join RB. 
Then AR — RB, [bij first pari of this theorem). 
To each of these equals add RQ. 
Then AR + RQ=RB + RQ. Ai. 2. 

But RB + RQ > qP. .Wiiy?) 

.-. \>v substitution, AR + RQ, or AQ > QB. Ax. 8. 

Q. E. D. 

113. Cor. Two points each equidistant from the ex- 
tremities of a line detertnine the perpendicular bisector of 
ihe line. 

This corollary gives a useful method of determining tlie 
perpendicular bisector of a given straight line, by determin- 
ing two points only of the perpendicular bisector. 



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I'lAKK LiEOMF/niY 



114. Dep. Tlio locus of a point is the pnlli of a point 
moviuL' a<!i;or(lin^ to ii given giiometrii; hiw. 

Tims, if a point more in a plane so as to be always two iiiehes Jia- 
tfliit from a given point, its tocua is tlie cirenraEereiiue of a circle 
wliose center is the given point, and wliose radius is & line two iuPhea 
in length. 

TliuB, also, the locus of a point moving so as to bo eqiiidisf!i,nt 
from two giren parallel liuos is a strait'lit line lyiug midway butween 
the two given lines. 

The locus of a point mny p,onKist of two or more sepa- 
rate lines or parts. 

Thus, the locus of a point moving so as to be always at a given 
distance from a f;iven line is two lines, one on either side of tlia 
given line, at the given distance from it. 

115. Demonstration of loci. In order to prove tlint a 
given line is the locns of a given point moving according 
to a given geometric law, it is necessary: 

1. To prove iJiai every 2>oiiit in the (jireii Hue i^irfisflrs tlie 
given hnr or condilion. 

2. To prox'c that every jwini not in the {/irrn line does not 
sdiisfif Ihe given law or comHtion. 

Instead ot 2, it may be proved tiiat every point which sjilisCes t.he 
given condition lies in the given line. 

Henee, in Prop, XX it has been proved that the per- 
pendicular bisector of a line is the locus of all points equi- 
distant from Die extremities of the line. 

116. Use of loci. Loci are useful in determining a 
point (or points) wLieh shall satisfy two or more geomet- 



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LOCI 53 

rieal conditions. For, by finding the loens of all points 
which satisfy one of^the given conditions, and also finding 
the locns of all points which satisfy a second condition, 
and then finding the intersection of these two loci, we 
obtain the point (or points) which satisfy both coaditious 
at the same time 

Thus, if it be required to find the poiuts wliich are two inubes 
from one glvon poiat and tbrne inchea from another given point, tbe 
two given points being four inubos apart, the required points are tbe 
intetsections of tbe circumferences of two t-irtles. 

Let the pupil inabo a construction and obtu,ia tho riiquiroJ point.H. 



Ei. 1. Dran 
inch from a piv 


f the locUB of rt point movi 
en point. 


iiK at the distal 


ice of 


one 


Ek. 2. Drav 
one inch from a 


7 exactly tbo locus of a poii 
, glvon line. 


at moving at a distanci 


9 of 


Ex. 3. Driv« 
distant from the 


r exactly the locus of a poii 
1 extremities of a givou line 


1 one iuohlong. 


..= .„,. 


Ex 4 n w 
tworaralicl 1 n 


man^ p i ts 1 1 -x plane , 
OS ' Ihieo panllel hneb ! 


(■^te irt 4- ) 


tule 1 




Ex 5 \r 
porrespouJ u„ 


two t 11 les Ljiil t th 
Ihr L ingk-* 1 tile tl r 


e ingl ^ of one 
t lHufttr.ite by 


1. al 
d iwii 


the 
ij, a 


Ex 6 1 a 
conne t the i ^ 


V th ee IS epbIob tiian ks on the same 
irti es Wb it truth is ilIustratLd 1 y th 3 


ijabe 
figure 


„d 



Ex 7 Draw a strait-ht 1 no ind locate a po i t " inches tii 
By the use of loei locate the po nts wh ch aie 1^^. incbes fi) 
given line iind aX the same diEtau>,o from the i, ven point 



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54 BOOK I. I'LANE GEOMETIIY 

Proposition XXI. Throrkm 

117. I. Evpry point in the iiseclor of ini angle 
diatani from the skies of the angle; aud 

II. Co^VERBViLV, every point equidixiatit from i 
of an angle lies in the bisector of the angle. 




I. Given TB the bisector of the iuieie ABC. P i 

poii5t hi PB, I'ij J. /:.L, iiiui FB ± no. 

PQ^FK. 

lu the rt. A PBQ aad PBB, 

PB^PB. Id 

ZPBQ=ZPBE. 1 

:. APBQ^APBR, Art, 

■c equal if the hypoteimse aii<l ait 



To prove 
Proof. 



Also 



{tu 



'- ofo 



Hence 



[honiolugotis siJes of equal &). 



II, Given ZABC, PQ1.AB. PE±BG, and PQ = Pli. 
To prove tlisit PB is the bisector of /.ABC. 
Proof. In the right & PBQ and PBS, 

PB^PB. Ident. 

Also PQ=PB. Hyp. 



yGoosle 



Or 



.-. A PBQ^-APBE, Art. 102. 

re equal if the hypolouise and a Icfj of onf, sU:. ) . 

:. Z AiiP= Z GBP (lio-mhgous A of equal A). 

ZABCh bisected bv BP. 

Q. E. D. 



118. Cor. In Prop. XSI it has been proved that the 
bisector of hh ntigk is the hens of all pohils fiquMistanl from 
the sides of the awjlr, for it h;i5 beon proved that every 
point in the given Hue siitisfioa tlio given law or condition, 
and that every point which satisfies thti given condition 
lies ill the giveLi line (soe Art. 115). 

E 

119. Dr.p. A transversal is a 
line that intersects two or more 
other lines. Thus, EF is a trans- 
versal of the I'nes AB and CD. 

If two lines are cut by a trans- 
versal, it is convenient to give the 
eight angles of intersection special 
names. 

n, h, 0, h, are called exterior ;uigles. 
c, d, e, /, are called interior angles, 
c, /, form a pair of alternate-interior angles. 
b, /, form a pair of exterior-interior angles on the 
same side of the transversal. 

Let the pupil name another pair of alternate-interior 
angles; also name another pair of exterior- interior 
angles on the same side of the transversal; also name a 
pair of interior angles on the same side of the trans- 

veraai. 




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56 iiOOK !. PLANE GKOMETRY 

PARALLEL LIKES 

120. Def. Parallel lines have already iieen ileiined 
(Art. 4:1) as stmiglit lines wliich lie ic the wimo piano iiiiil do 
not meet, howcvci' far thoy bu produceil. 

What is the fuiic!ameutal axiom eonccrniiig pafallel 
lines? (see Art. 47.) 

PROrOSITION XXII. TUF-ORKM 

121. Two straight lines in Ihesatne plane, perpendicular 
to the same straight iiite, are parallel. 



Given the lines AB and CD J. line FQ i\iid in the 
same plane. 

To prove AF, || CD. 

Proof. If AB and CD are oot parallel, tliey will meet 
if sufficiently prodiieod. Art. 120. 

We shall then have two Js frura the same point to the 
line PQ. 

But this is ii 



Heace AB and CD never meet. 

,■, AB and GD are parallel. 



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PAltALLEL LINES 



57 



122. Cob. Twosiraight lines parallel to a third straight 
line are parallel to each other; 

JAnes parallel to parallel lines are parallel; 

Lines perpendicular to parallel lines are parallel; 

Lines perpendicular to noii-paruUel lines are not 2>arallet. 

Proposition XXIII. Theorkm 
123. If a alraii/M line is perpendicular lo mir of iu:o 
given paraUcl lines it is perpendicular to the other also. 



Given An II CD, i\w\ PQ L AB. 

To prove PQ 1 CD. 

Proof. Let or be drawn X FQ at 0. 

Then CFWAB. Art. m. 

IfKo siraigltl lines in the savic plane X snmr. sirnUjhl liiw are ||). 



But 



{through n. giccn i<oi 



CD II AB. 

CF coincides with CD, 
I one straight- line, aiul only i 

aiioOicr gir':a straight iinv). 



Bnt 
Hence 



Hyp. 



{for CI) coincides with CF, to ii-hich FQ is X)- 



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DO BOOK !. rLANV; Ci;(>?ilETItY 

Proposition XXIV. Tukokiiji 

124. // I'ro parallel sirnigM Uihx irrr ■■nt hij a trans- 
tersal, llw allcruate inlcrwr <iiighs '.irc equal. 




Given the 11 lines AB and C7' (;ut by the tt-aiisvevKai 
I'Q at the points G and E respectively. 

To prove lAGE^ KjED. 

Proof. Thi-ongli R, the middle point of EG, let the line 
ffFbe drawn J. AB. 

Then HF X 6'/), Art. 1^3. 

Ufa straight line is X one of tico i] (iiiw, il is X the oiiur alsii). 

, 111 the right A QJtH and EEF, 

GE^EB, Consw. 

/tOMS^-ZEBF. (Whjl) 

.-. A (Jffiff = A ERF, Art. 98. 

(too right A (wa ejwa^ ^ the hypotenuse and an aauie Z of one =: 
(Ae hffpot. and an acute / of Vie other). 

:. I HGR = Z REF, or, IA0E= I OED, 
Uiomologous A of equal A). 

Q. E. B. 

Es. In tte above figure let the pupi! show that Z BGE = Z GBC. 



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PARAiLEL LINES O'J 

Prop. XXV. Theorem (Conveese op Prop. XXIV) 

125. If two simujhi lines are cut hij a transverml, 
making the aUenuite interior angles equal, the two straight 
lines are parallel. 




^Given tlie t'wo lines AB and CD cut by tlie triinsveraal 
PQ at the points F and ff, making Z AFG - I FGI>. 
To prove AB \\ CI). 

Proof. Throngh F let the line KL be dmvvn li CI). 



Then 

{iftivo II si. 

Bnt 


. lines 


I KFG = Z FGl), 

 are ciU hy a transversal, the all. int. S a 
IAF6 = IFGD. 


.\i-t. 124. 
rv equal). 
Hyp. 


Hence 




IKFQ=- I AFG. 
:. KL coincides with AB. 


As. 1. 


Bnt 
Heuce 


(fo> 


KL II CD. 
AB 11 CD,  
 Al! •:oincides milt KL, wlurh is \[ CD). 


Conatr. 



Ex. 1 . Tf Z BFG = Z FG C, prove that AB aid CD a 

Ex, 2. By exact 
use o£ rulai' and 
eompasaes , lit a 
given point {P) iu 
agivenstraigbtline 
{OA) poivstiiiot an angle equal to a given Z(B), 



Ex. ; 



lelti 



lethods, tbrougti a giv 



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PitOPOSiTiox XXTI. Theoieem 
126. 7/ ta-oparnlM lint^s are cut bij a transversal, the 
exterior interior (Niy/i-s tire eqiuil. 



Given the || lines AB untl CI) cut by Uie transvprsa! PQ 
at the points Fand G respectivelj-. 
To prove ZPFB^ZFOD. 

Proof. Z FFB = Z AFG. (wiiy! ) 

IFGD= Z ,-lF(7, Art. 121. 

Cbeiitg all, uii. A of inirnlld Hiics). 

.-. IPFB^Z FGI>. As. 1. 

lu like maniipi- it may be shown that AFFA - Z FGC. 
q. E. B. 

Prop. XXVIL Toeori^m (t'oxvEKSK ov Prop. XXVI) 

127. If Uvo HtraUjlii linn arc ad J»j n irani:rn-Kal, nuik- 
ing the exterior interior angles equal, the two siniiqhi lines 
are parallel. 

Given, on Fig. of Prop. XXV, ZPFB = IFGD. 

To prove AB 11 CD. 

Let the pupil supply the proof. 



Ex. If, ill l^lie Fig. to Prop. XXVI, /. ['FB equaU €.7°, Ciad tho othi.1 



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PAEALLEL LINES Gl 

Proposition XXVIII. Theorem 

128. If two parallel lines are cut hy a transversal, the 
sum of the interior angles oii the same side of the trans- 
versal is equal to two right angles. 



Given the ll lines AB and GH ent by the tmnsversal PQ 
at tlie points F and 6 respectively. 

To prove Z BFO + IFGI)=2 vi. A . 
Proof. /iFGI> = ^FFIi. Art. ISO. 

To VM-], of tliese eciiials add Z BF(i. 
Then / BFO + IFGD= IPFB + Z BFO. Ak. 2. 

But Z FFB + Z JiFO^-1 rt. d . A,t, v.i. 

.-. IBFG + -^FOB^^rt. A. Ax. i. 

Q. £. D, 

Pkoi>, XXIX. Thf.(irf.m (CotiVERSF; oe Prop, XXVITl) 

129. // iiro straight lines are- nit hy a transversal, 
maMng the stun of the interior angles on the same side of 
the trai»iversal hiuuI to tieo right augles, the dm lines are 
varalkl. 

Given, on F\<r. of Prop, XXV, Z BFG + Z F0JJ = 2 rt. A . 

To prove AB \\ CD. 

Let the pupil supply the proof. 

Ex. If, ou Fig. of Proi.- XXVIil, ^i'Fli + ^QGD^IBO", are Jli 

«d CD II r 



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? BOOK I. PLANE GEOMETKY 

Proposition XXX. THrcoKicjr 

130. Two angles whose sides are paralld, vach to each, 
•e either equal or supplei. 




Given A7i || JiR, and EC II GF. 

To prove A.ABC, DBF and GKIT pqiial, and AABC. 
and DEG supplementary. 

Proof. Produce the lines JiC and Hl^ to interseet in P. 

Then Z ABC = I QPE, and Z QFE = Z DEF, Art. 120. 
(fceinp eri;i. int. A of || lines). 

.-. IABG= IDEF. (Wliyf) 

Also ZDEF= ZGER. CWliy!) 

.-. lABC^ IGEH. (.WhyO 

Ajjain .i Ti/JF and BEG are siipplementaiy. Art. 73. 

.-. A ABC and DEfJ are snppleraentai'v. Ak. 8. 

Q. E. B. 

131. Note. It is to be observed tlmt in tbe above theorem the 
two angles are equai if, \a the pairs o£ parallel sides, hoth pairs estend 
in the same direotiou Irom tiie vertices ( i B and DEF), or both paira in 
opposite directions ( i B aod GEH) i and that they are supplementary it 
tmepair extends in the same direction, aud the other pair in opposite 
directions { ii B and DEG). The directions ot the sides are deter- 
mined by connecting tlie vertices of the angles and observing wliether 
the liaes considered lie on the same side or on opposite sides of the 
line drawn. 



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PARALLEL LINES 



PsoposmoN XXXI. Theorem 



132. Two anglfK ; 
each are either equal a 



hose sides are perpendicular each to 
supplementary. 



Given BA ± EF, and nC X TiD'. 
To prove ^J.J5C'= IFEI), ami A ABC and FED' sup- 
plementary. 

Proof. At E let line EK be drawn ± EF and in tlie 
same direction with BA ; also EG L !>!>' and in the same 
direction with BG. 

Then BA 11 EK, and JJC 11 EG, AH. I2i. 

(iji'o sirciiijht lilies in tlie same plane, J. (lie same straight line, are H), 

.-. LABG = /LEEG, Arts, m, ni. 

((wo ^ whose sides are ||, eacft to each, aiid extend in the same lUreelii/n 

from the vertices are =). 

But I KEG is eoraplement of IBEK, Art. .t^. 

{for IDEGisart. I lif coiislr.) . 
Also Z.FET> ia eoniplement of IDEK, Ait. 33. 

{for I FEE is art. L by conslr.). 

:. IFED^IKEQ, Art. T5. 

{,eoT>ipleme»ta of the same L are = ). 

.-. LABO=LFEJ). Ax. 3. 

But I FED' is supplement of / FED. Avt, :i4. 

.-. ZFi'i'' is aupplement of ZAiiC. Ax. 8. 

Q. E. D, 
133. Note. In the above tteorem the two angles ore equal if the 
sides, oonsidert'd as rot.iting about tiio vertices, are taken in the same 
order (thus SG ia to the right of BA, and ED to the right of A'."' 
■'■Z.ABC= ^FEV); but tlie angles are Bupplementary if theeorre'i- 
ponding Bides are tiik-flii in tlio opposite onltr (llins,^Cis to the rii:lit 
ttSAXKHEiy is to tku left of EF :. Z JZ)'C' = Buppleineiit et IFti:!'). 



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PROrOSITIOX XXXII. TuEOTiKJI 

134. The Slim, of the. unyles of a Irian.,},' h i'qmil io tiv 
rUjhi migli's. 



Given tlifi A A BC. 

To prove lA+ZIi + l nCA - 2 vt. A . 
Proof. Pi-odnfie the siile AC totbe point. 7) sniil theon^h 
C let the line CE be dmwn || AB. 

Then Z Kr7) + Z iSCB + Z .fiCA = 2 rt. ^ , Art. 77. 
fi/ic siiiH i)/n!i llie A alHiiit a point on the same side of a su-tiiijlii line 
pii.'isiiiij tliiojigh the poi>it = 2 rt. S. ). 

But IBGB^LA, Art. 120. 

{fe/Hfif (xi. ini. ti. ofpnruUel U»es). 

Also IBCE = ZR, Art. V2i. 

{hil,u, nli.int. A ofp'ir„!lfUiw,i). 

tSiihslitiitiiigfi.r ZICCD itseiiual, /.4,ai)d fnr IBCE 
it.^ equal, Zli, Ax. 8. 

/A + zn + ZnVA^'l vt. A. 

0- E. D. 

135. Cor. 1. An exterior angle of a triungU is eqwil to 
the sum of the fico opposite interior angles. 

136. Cor. 2. The sum of any two angles of a triangle 
is less than two right angles. 

187. CoE. 3. In a right triangle the sum of the two 
acute angles equals one ri'jht angle. 



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PARALLEL LIN'ES 65 

IS8. Cor. i. A triangle can have but one rtgM. or one 
obtuse angle. 

139. Cor. r>. If two angles of one triangle equal two 
angles of anolher tnangJe, the third angle of the first tri- 
angle equals the third angle of the second. 

140. CoE. 6. If an acute angle of one riyht triangle 
equals an acute angle of anotlmr right triangle, the rentain- 
ing acute angles of the triangles are equal. 

141. Cor. 7. Tiro triangles are equal if two angles and 
a side of one are equal to two angles aiul the homologous side 
of the second. 

142. GoR. 8. Two right triangles are equal if a leg and 
an acute angle of wie are equal to a leg and the Iwmologous 
acute angle of the other. 



Ex. 1. If two nngips of tt triiinelG sire M° ami G'i^", find the re- 
maining angle. 

Ex.2. If one nciito .nngle of n, riglil trinngle is r>fi° 1"/, find Ihe 
other acute niiglc, 

Ex. 3. How xnKny lifgi'Gos in each angle of an equilateval triangle? 

Ex. 4. IIow nianv degrees in eatli acute angle of an isosceles 
nsht liianslo! 

62", 72°! 

Ex. 6. If one angle of a triangle is 4^'°, find the sum or the olher 
two angles. 

Ex. 7. If two nngles of a triangle are 3S° and 65°. find all the 
est«rior angles of the triangle. 

Ex. 8. If tlu^ verti's aiifjie of an isoiicelea trinngls is 3S°, fiml eaeli 

Ex.9, If an angle at the Use of an i>,llSl^eios triangle is .'iO", find 
tbe vertes ani;li-, 

Ex. 10. All f\terior aiisle at tlie liasu of an isosceles triangle is 



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I'LANK <;i:oiiini(Y 



QUADRILATERALS 

143. A quadrilateral iri a portion of a piano boumlod. 
by four straiglit lines. 

The sides of a quadrilateral are the Ijounding lines; the 
angles are the angles made by the bounding lines; the 
vertices are the vertices of the angles of the quadrilateral. 

The perimeter of a quadrilateral is the snm oi the sides. 

144. A diagonal is a straight line joining two vertices 
that are not ailjacent. 

145. A trapezium is a quadrilaieral no lno o( wlioso 
sides are parallel. 

146. A trapezoid is a quadrilateral whicli lias two, and 
only two, ot its sides [Kiiallcl. 



147. A parallelogram i; 
sides are parallel. 



a quadrilateral whose opposite 




14S. A rhomboid 
oblique angles. 

149. A rhombus is a rhomboid whose sides are equal. 

150. A rectangle is a parallelogram whose angles are 
right angles. g' 

151. A square is a rectangle whose sides are equal. 



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QUADEILATEE.U^S 

152. The base of a parallelo- c 
gram is tlie side upon which it is / 
supposed to stand, as AB. The / 
opposite side is called the upper "* 
base (CD). 

The altitude of a parallelogram is the perpendicular dia- 
tance between the buses, as JHF. 

153. The bases of a trapezoid are its two parallel sides. 
The legs of a trapuzoid are the sides which are not parallel. 
The altitude of a trapezoid is the perpendicular distance he- 
tweeii the bases. The median o£ a trapezoid is the line join- 
ing the midpoiuts of the leg:-. 

154. An isosceles trapezoid is a trapezoid whose legs 



Ex. 1. Draw a quiidriltiteval with three acute 

Ex. 2. Is every rhombus a rhnuitioid J Is 
thombus ? 



Bx. 3. \Thatis the ilifferenee b( 
What [iropeilies do they have 

Ex. 4. Find the perimeter o£ a square foot 
Bj Ibid of the following clflBSificatfoii: 



glee and one ob- 

^ery rhomboid a 

iqwui'e and a vhombuef 



Quadrilateral 



^ Trape^^oid , 
I. Pavallelogr 



. Isosceles trapez 






) Rhomboid . . . rhombus. 

Ex. B. Determine what four mimes the rhombus is entitled to. 

Ex. 6. Datevmiiie what praporties Ihe rbombua, square and 
rectangle have in commou, 

Ex. 7. A diagoniil of a rhombus divides tha rhombus into how 
JUany tviaiiglea F What kiud ot triangles are these {  



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BOOK I. I'LASE (;i:OMETliY 



Propositiom XSXIII. Theoreji 

1 55. The opposite sidi-s of a parallelof/ram are equal, and 
its opposite an'jtfs (ire also equal. 



Given the paralklograni ABCI). 
To prove AJ)=BC, AB^'DC, in=^D, aud lBAJ>-= 
I BCD. 

Proof. Draw the diagonal AC. 
Then, in the A ABC and ADC> 

AC^AG, (Why?) 

IBCA = ZCAD, Art. ui. 

{being alt. int. i of parallel lines). 
ABAG= AACD, 
(some reason). 
:. £^ ABG = C^ AGB, Art. nr. 

((H'O A are cqwil if ia-ii A and ihn iiielii/lcd Kide of one an equal 
rc.iperliri'ly U> liio A aii'l the. incbidvl side of the other). 
:. AD^BC. A «= BC and Z B= Z 1), 
ih,moloyo„s pans of equal &). 
Ill like maimer, hv drawing the diagonal BD, it may be 
proved that IBAB^IBCD. ^ ^ ^ 

156. Cor. 1. A diagonal divides a parallelofjrain into 
two equal triangles. 

157. Cor. 2. ParallelUnes comprehended between par- 
allel line.i are equal. 

158. GoR. 3. Two parallel Uhes are everywhere equi- 
distant. ^ , 

Ex. 1. In the above figure, prove ^SJO=ZflCD byuae of Ax. 2. 
Ejc. 2. Prove tiie oppoeite angles of a, p stall el og ram equal, by use 
0£ Art. 130. 



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quadeilateeals; by 

pROPOSmoN XXXIV. Theorem 

159. If the opposite sides of a quadrilateral are equal, 
thfi figure is a parallelogram. 



Given the quiuli-il^t.eral ABCT) i 
F.€=AT>. 



To prove ABCl) a dl . 




Proof. Dra-sv the diagonal AG. 




Then, in the & ABC and ADC, 




AC=AG. 


(Why ?) 


BC^AT). 


(Why?) 


AB=CD. 


(Why 5) 


.: AABG^AADC. 


(Why ?) 


.: Z E.-1C= Z AGT>. 


(WhyT) 


:. AB II CJ>, 

• (ICO lines are ,:i<t h,/ i, Irami-crsal, miihii'i the alt. i 
IhKl: arc |j). 


Art. 125. 
III. i. equal, the 


Also Z£eA = I CAD. 


(Why?) 


:. BG II AD. 


Art. 12,-K 



.'. ABOI) is a C^, 

a ijHUitrilaleral tcliusv ojipo. 



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70 iXlOli; I. PLANE OEOlIETliY 

PiiOPOsiTioN XXXV, Theorem 

160. // tivo sides of a qiimh-ilateral are equal and par- 
allel, the other tiro sides are equal andpuraUcl and the figure 
is a paralletoyrain. 



Given tlie qiiaaiitaterni A liCD in wliieli BG = and ); AIJ. 
To prove AB(JD a d/ . 



Proof. Draw the diagonal .4 C. 






Then, in U 


ic A ABC ami ADC, 








AC=AC. 




(Why!) 




BC^AB. 




(Wby 1) 




IBCA= ICAT). 
{bchig •ill. u,l. A of jMuilM Unr^) 




Art. 124. 




:. A ABC^AABC. 




(Why?) 




.: I BAC=IACB. 




(Wliy t) 


tifo Juics arc 


:. AB\\ CB, 
Ml by a lnii::<i-fT.i(\l, mildinj Ike all 
Iwes are [[) . 


. /(If. A 


Art. 125. 

(■.,"ll/, Ilw 




.-. ABCI) \& 3. rj . 


■nt augl. 


Ai't. 14V 
J. E. D. 


Ex, 1. Shon 


 that in a Z35 each pair of adjai:e 





plementary. 

Ex. 2. One angle of a parallelogram is 43° ; find the ottier an 
Ex. 3. If, in the triangle ABC, Z J=60°, Z2i = 70°, whicli i; 

longBRt Bide in the triangle I Which the ehorteat ! 



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QUADRILATKllAI-S 71 

Proposition XXSVI. Theorem 
161. The diagonals of a parallelogram bisect each other. 




Given the diagouuls AC aiul Bl> of the £UAHC'D, 
intersecting at F. 

To prove AF=FG, and r.F=FTi. 

Proof. Let the pupil supply the proof, 

[SuG. In the A BFC and AFD what si.les uro equal, aiul wby ? 
What A are equal, and why 1 ett.] 



£eure V 

Ex. 2. If one aiiKie III n parallelogram i^ tlu-ee (iiiiL-saQotliev angle, 
find ali the angles of l!ie parallelogram, 

Ex. 3. If two angles of a triangle ace j." ami v°, find the third 
angle. 

Ex. 4. If two angles of a triangle are j" and OU^ + j°, find the 
third angle. 

Ex. 6. If OBB angle of a parallelogram is a°, Ihid thu other uncles. 
Ex. 6. Construct exactly an angle of W. 

Ex. 8. liow hirfjB way the doutile of an ubtuab uuyie be ! how 



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l-i 1500K I. I'LAXi; GKOMiyrRV 

Piim*osrrio.N' XXX VI i. TiiEdiii-.M 

162. Two parnJMogramH ui-'' eqind if two ailjticcnt sUhs 
and tJie included angle of one are, cqii/il, rcspuctinhj, io l/ro 
adjacent aides and the inchidvd a>i;/l(.' •■fihi' uthn-. 



Given ihe CD ABV}^i\m\ A'li'Ciy uwyVwh A ti = A' IV, 
AD = A'I)', and lA =IA'. 

To prove LTJ ABCD = CI7 A'B'C'D'. 

Proof. Apply the £17 A'B'G'D' to the CJ ABCT) so that 
A'D' shall coiueide with its equal AD. 

Then A'B' will take the direction of AB (for IA'=IA); 
•and point B' will fall on B (for a'B'=ab). 

Then B'C and BC will both be 1 1 .-ID and will both 
pass through the point B. 

:. B'C will take the direction nf BC, Geora. Ax. .■!. 
(thro)igh a gircn j>oiiit one siraighi line, ami oiilij one, can he {Irairii \\ 
aiiolktr gh'cn stmiglit liii':}. 

In like manner, !>'€' must take the direction of />('. 

.-. C must fall on C, Avt. C4. 

(tuto siraighi lines can iittersci-t in hiil one Jioint]. 

.: Cn ABCI) = CD A'B'diy, Art. 47. 

(geometric figures it-Mch eoiiicitie are eqval). 

Q. E. D. 

163. Cor. Two rectangles which have equal bases and 
equal altitudes are equal. 

Ez. Construct exactly an angle of 30°, 



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164. A polygon is a portion of a [<lane boutitled by 
Straight Hues, as ABCDE. 

The sides of a polygon are its bounding lines; tlio 
perimeter of a polygon is the sum of its sides; the angles 
of a polygon are the angles formed by its sides; the 
vertices of a polygon are the vertices of its angles. 

A diagonal of a i)o!ygoti is a straight line joining two 
vertices whkdi are not adjacent, aw IIT> in ¥\^,. 1. 





165. An equilateral polygon is a jiolygon all of wliOKe 
sides are equal, 

166. An equiangular polygon is a polygon all of whose 
angles are equal. 

What four-sided polygon is equilateral but not eqiii- 
angular ? Also, what four-sided polygon is both eituilateral 
and equiaugular ", 

167. A convex polygon is a polygon in which no side, 
if produced, will enter the polygon, as ABODE {Fig. 1). 

Each angle of a convex polygon is less than two right 
angles and is called a salient angle. 

168. A concave polygon id a polygon in which two or 
move sides, if prudiK^ed, will enter the pob^gou, as FGUIJK 
(Kg. 2). 



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/4 HOOK T. PLAXE r.EOilEI-llV 

Some angle of a ooneave polygon must be greater than 
two right angles, as angle GlII of Fig. '2. Such an angle 
is termed a re-entrant angle. 

If the kind of polygon is not speui&ed in this respect, a 
convex polygon is meant. 

169. Two mutually equiangular polygons are polygons 
wliorii- eoi't-espoiiding angles are equal, as Figs. 3 and 4. 



170. Two mutually equilateral polygons are polygons 
■whose corresponding aides ai'e equal, as Figs. 5 and 6. 

From Figa. 3 and 4 it is seen that two polygons may be 
mutually equiangular ■without being imituaily equilateral. 
What similar truth may be inferred from Figs. 5 and 6f 

171. Names of particular polygons. Some polygons are 
used 50 frequently that special names have been given to 
them. A polygon of three sides is called a triangle ; one of 
four sides, a quadrilateral ; one of five sides, a pentagon ; of 
six sides, a hexagon; of seven sides, a heptagon; of eight 
fiides, an octagon ; of ten sides, a decagon ; of twelve aides, 
;i dodecagon ; of fifteen sides, a pentedecagon ; of » sides, an 
n-gon. 

Ex, 1, Let the pupil illustrate Arts. 169 and 170 by drawing two 
pentagons that are mutually equilateral without being mutually equi- 
aDgalar, and another pair ot which the reverse is true. 

Ex. 2. Can two triangles he mutually equilateral without being 
mutually oquianguiar ? What polygons can f 

Ex. 3. How does the iiumbet of vertices in a polygon uompare 
witlj the number of aides ! 



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polygons '■> 

Proposition XSXVIII. Theobrm 

172, The sum of the angles of any polygon is equal to 
iipo right angles taken as many times, less two. as the poly- 
gon has sides. 




Given a polj-gon of n sides {the above polygons of 5, 
6, 7 sides being used merely as particulai- illustrations, to 
aid in carryicg forward the proof) . 

To prove the sura of its A = (ji— 2) 2 rt. A. 

Proof. By drawing diagonals from one of its vertices 
the polygon is divided into {n — 2) triangles. 

Then the sum of the A of each triangle = 2 rt. /^ . Art,i34, 
(lite mm of the A of a A w eqmt to 3 rl. A ). 

Heiieethe sum of the Aot the {h— 2)&= (n — 2) 2 rt. .£ . 

Ax. 4. 

But the sum of the A of the polygon is equal to the 

sum of the A of the {ii — 2)^. X-a. S. 

Ileaee the sum of the A of the polygon = {« — 2) 2 rt. ^ . 



173. Cor. 1. Thi> sum of the angles of a polygon equals 
in— 4) rt. A. 



eqiiiangidar polygon of n 
rt. A. 



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7() ]iOOK I. PLANK GEOMl/rnV 

Pkoi'OSITIOX XXXIX. TuF.oiiKM 

175. The sum of llif cxlerim- idi'jU-s of h pohjijoii formeO 

hif pro'liMinil it.< siik.'i in ^sun-r.-'.'i'ii/ ill cif i:.flri:iiii!ij fijiitii.- 



four right a>Hjk, 




Given a polygon of n sides hiivlng iis buIcs pi-oiliiced 
in suceession. 

To prove the sum of the exterior A =4 rt, A . 
Proof. If the iiiterioi" A of the polygon he denotod by 
A, B, C and the corresponding exterior A by «, h, <\ 

lA-^ lfi^2 rt. A, (VThj!) 

l!i + l}) = 2 rt. A , C^'hy ?) 

etc. 

Adding, int. A +ext. A = w time.'; 2 rt. :i =2jM-t. ^ . Ax. 2. 

But int.i = ((/— 2)2rt.:i=2» vi.A — \vi.A. Art. I7:i. 

.-. Ext. i=4 rt, A. 

Q. E. D. 

El. 1. What does the aum of the interior angles of a lieaagon 
equal ? of a heptagon t of a decagon '. 

Ex. 2. Each angie of an equiangular pentagon eontaina how many 
degrees ? of au equiangular hexagon ! octagon ? decagon ! 

Ex. 3. Would a quadrilateral constructed ot rods hinged tit the 
ends (i. e., at the verticea of the quadrilateral) be rigid ? Would e, 
triangle so constructed be rigid J Would a pentagon ( 



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MISCELLANEOUS THE0EEM8 



MISCELLAKEOUS THEOREMS 

Proposition XL. Theorem 

176. If three or more parallels intercept equal parts on 
owe transversal, they intercept equal parts on every trans- 
versal. 




Given AP, P,Q, OR nnd T>T parallel lines intercepting 
equal parts AB, BC and CD on the transversa! AB. 

To prove that thev intercept equal parts PQ, QR and 
liT on the transversal FT. 

Proof. Throngh A, B and C let AF, BO and CR be 
drawn parallel to PT and meeting the hues BQ, CR and DT 
in the points F, 6 and H respectively. 



Then 

{twos. 
In the 
/ CDI{, 

Also 
Aut 



But 
{parallel ii 



the lines AF, BO and CE are li , Art. 122. 
aislU hues II a third straight line are || eai:h vther). 
%. ABF, BCG and CDM, Z ABF = Z BCG = 
{being ci-i. int. A ti/|| Unm). 
IBAF^ ICBQ= IDCH, [Mmcreason). 

AB=BC'=CD. Hyp. 

.-. AABF=A BCG^ACBH. Art. 97. 

.-. AF=BG = GR. (Whyn 

AF^I'Q. liG=QR, CU = RT, Art. 157. 

ifs 0("i;ii-f/(n(i;«[ bc(ii'p«i. pnraW lines arc equal). 

.-, PQ^QR^RT. Ax. I. 



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PnOPOSlTIOX XIJ. TllEdRElt 

177. The line xvUdi join^ flw mhlpnhiU of livo shUs of 
a triangle is parallel to the third nidu, and is equal io one- 
half the third side. 




Given J) the midpoint of AB, and J-J the midpoint < 
AC in the triangle ABC. 

To prove BE \\ BG and =iBC. 

Proof. Through B let BL bo drawn 11 AC, and meetir 
DE produced at L. 



Then, in the A ADE and BDL, 




AD^BD, 


(Why!) 


lABE^ZBBL. 


(Wliyr) 


ZDAE^ZDBL. 


(Why?) 


.: AABE^ABOL. 


(Why?) 


.: I'K^JIL, or DE^h BE, and A l-]---BL. 


(Why?) 


Bnt, A E ^EG {llyi^.) :. FA'.^-HL. 


Ax. 1. 


Al«o EC 11 BL. 


Constr. 


:. BLEC is ;i CJ , 


Art. 160. 


' Ucu Hides of a quiulrUuleral arc equal ami ptirtdlH the figio 


■eisfl^Z?). 


:. BE 11 BC. 


Art. 147. 


Also LE^BC, {opp. sides ofaCy are =). 


Art. 155. 


.■.iLE,OTBE = iBG. 


As. 5. 




q. E. I). 



178. Cor. The line which hiseds one side of a triangle 
and is parallel to another side bisects the third side. 

Thus, given AD = DB and DE 11 BG, then will AE=E<J. 
For, suppose a line FG drawn through A \\ BG, 



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MISCELLA.MEOUS THEOIUIMS 



79 



Then the three parallels FG, DE, JiC will intercept 

equal pai'ts on AB. Hyp. 

,". they intercept equal parts on AC. Art. 1T6. 
.-. AE^EG. 

Proposition XLIT. Tiieckem 

179. The line wkieh joins the midpoints of the legs of 
a trapezoid is parallel to the bases ami equal to one-half 
their smn. 



Given the trapeKoiil ABCB, E the mklpoint, of the leg 
AB, and F t\\e midpoint of the leg' CB. 

To prove that a line joining E .iiul /'Is 1| AT) and BC 
and = i UD+ BC). 

Proof. Draw the diagonal BI) and take (r its mid- 
point. Draw EG and GF. 

Then, in" the A AfW, BG\\ABa\i(\ = A AT). Art. J77. 

{tU line which Joino f!ic miilpnfiits nf dm ^iili-^ iif a i la [l Ihe (liwd stile 

and = oiic-half tin: third eiile). 

Also, iutlie Ai'i^C, GFllliCand^i BC, {sawc r.r^mn). 

.: OF ami AD hoihwnC; .-.CFwAI), Art. 123. 

{two striiii/Jii lines \\ a t/iirU stnUnht are \l each oi/iei-): 

.-. EG and GF are both 1| AD. 

• '. EG and GF form one and the some straight line EF, 



( Oirmigk a giccn point one line, and onhj one, caa be drawi 
gixcnImcS. 
:. EF\\ABm(kBC. 
Also EG = I AB, and GF=l BC. 

Adding, EG + GF, or EF=^ {AB^BC). 



180. Cor. A line draicn hi.'fecfh: 
and parallel lo fhe base hisccln the oh 



OilclPf, of, 
r hij also. 



mioliier 



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]iOOt; I. rj-ANE GEOJIKIUY 



Propositios XLIII. TuEOiiEir 
Tlu- Inscrtors of thr am/kn 'fa trlimt/ir 




Given Uie A ABC with tlic lines .1/', BQ, CJi (Fig- 1) 
bisecting the/. BAC, ABC, ACB, respectively. 

To prove tlmt AP, BQ, CR intersect in a common point. 

Proof. Let AP, the bisector of / BAC, aiul CR, tlie 
bisector of / BCA, intersect at the point 0. 

Then 0, being in bisector /IP, is equidistant from AB 
and AC, Art. 117. 

(every point in the hmctor of nnLU equidistant from Hie tides of IheZ). 

Also 0, being in bisector CR, is equidistimt from .If" nncT 
Ii( ', {mini! iT'fXoii). 

llenuc O h cqiiiOistaiit I'roni tlie riidus ,-l/>' anJ li< '. Ax. 1, 

. -. (I is in the lilsector of /. A BC, Art. i it. 

{every point e^idiliMiintfrom the sides of'luZ lies in the hiMVtor of Qia Z.) . 

Hence BQ, or BQ prodtiueil, passes through 0. 

Hence the biseetors, AP, BQ, CR, of the three A of the 
A ABC intersect at 0. Q, E. D. 

182. OoR. The point in which the three bisectors of the 
angles of a triangle intersect is equidistant from the three sides 
of ike triangle. 

1 83. Dbf, Concurrent lines are lines which pass through 
the same jioint. _„_____ 

Ex. Find other Z of above flgurs if Z /M f = 73= aud Z BCA =44'. 



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MlSCELLANEOrS THEOREMS 



Proposition XLIV. T 

184. The perpe-ndicular bisectors of the sides of a trian- 
gle intersect at a common point {called the circumcenter). 




Given tlie A ABi- with DP, EQ, Fli. (Fig, 1i, Uio j_ 
biseclufa of tliu sides AB, H<.\ C.\, reKpectivoIy. 

To prove tiiat T)P, EQ, FR intersect at a comimin point. 

Proof. Let DP and EQ intersect at the point 0. 

Then 0, being in _L OP, is eqnidistiiut from tlie jjoiots A 
and B, Art, 113. 

{ever-ff poiiU iii, We perjieiulkiilttr bisector of a line in equaily dieivnifrmn 
ilte exbremiiiee of ike Urie) . 

Also 0, being in J_ EQ, is eqnidistunt frorii the points 
B and C, 

(mme reawu). 

Hence O is cquidist[uit fram .1 imd ( '. Ax. 1. 

.-. is in the ± bisector of AC, Ait. 115. 

{the, \_ hiaector of u line U Hie locus of alt points equidisUfiil from the 
exireifiitiei! of the line). 

Henee FR, or FR prodnoed, passes throngli (). 

Hence the perpendicuhir bisectors, }iP, EQ. FR. <i\' ibc 
lliree sides of the A ABC meet in tlio point 0. a. e. a. 

185. CoJ!. '!'}i<- pi)i))l ill which lin; pi'r}n'iidivuhir hiKi'vlKn^ 
of Ih,; .u(l,s of a iriuiujl, mcvl is equidisloiit from thi- verliim 
of the triaiujU: 



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[■LANK GEOMF-TRY 



rilOroSITION' XLV. THEORE^il 

186. Tlif perpi'mVculars from the ret-firca of a triangle 
to the oppogi/e sldca iiuet ut a point {called Ihe ortho-center). 





'A 


N 




M 


/^ 


w 





Given AD, BF, am\ CB the perpendiculars from the ver- 
tices A, B, and C of the A ABC to the opposite sides. 

To prove Ihiit AD, BF, and CE iotorsect in ;i comiiiou 
point. 

Proof. Through A, B, and C let the lines PR, VQ and 
QR be drawn || BC, AC, and AB, respective! y, and forming 
the A FQR. 

Then AD 1- VB, Art. 123. 

(/,»■ AD 1 BC. njjrf r, U:ir J. n:ii; of lli-o [[ !i»rs U 1 the Oilier ulm) . 

Also A ]' BO imA A BCli are Z17 . Constr. 

.-. AT = BC, and AB = EC. Art. 155. 

(((((! opponiic si(}es of a CD are = ). 

.-. AP = AR. As. 1. 

lience, in the A FQR, AD is the perpeudicukr bisector 
of side PR. 

In like manner it may be shown that BF is the perpen- 
dicular bisector of FQ, and that GE is the perpendicular 
bisector of QR. 

Hence AD, BF, and CB are the perpendicular bisectors 
of the sides of the A PQB. 

.: AD, BF, and CE meet in a conimon point, Art. 184, 
ithi }>ci-pendieula,r biseeiois of the sides of a A arc concurrent). 
Q- E. B. 



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miscellaneous theokems 83 

Proposition XLVI. Theorem 

187. The medians of a triangle intersect {or are con' 
current) in a point {called i?ie centroid) which cuts off two- 
thirds of each median frotn its vertex. 




Given AD, BF and CB the three medians of A ABC. 
To prove that AD, BF and CB intersect at a point which 
cuts off two-thirds of each median from its vertex. 
Proof. Let the medians AD and CE intersect at 0. 
Take R the midpoint of AO, and S the midpoint of OC, 
and draw RE, ED, DS, and SR. 

Then ED 11 AC and = I AC. Art. 17T. 

Also, in the A AOG, 

lis II AG and - i AC. Art, 177. 

ED II RS and = ie^'. Art. 122 aad Ax. 1. 

.-. REDS is a £3' . 

ES and RD bisect each other. 

AR^BO, and CS=^SO. 

.: AR = RO=OD, and CS^^SO^OE. Ak. i. 

Hence t'£erosses.'IJ.'at a point 0, such that A0=|.4i). 

In like manner it may be shown that BF crosses AD at 

the point 0. 

Uuuite the medians A D, BF, iuid OF intersect at point 0, 
which cut-s otf two -thirds of each median from its vertex. 

Q. E. D. 



Hence 



Hence 
■Bnt 



(Why!) 
(Why!) 



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84 



BOOK I. TLAKE Cr.OHETEY 



188. Properties of rectilinear figures to be proved by the 
pupil. Proof of equality of triangles. Otlier properties of 
reetUioear flgares will now be given vvliioh are to be demon- 
strated by tlie pupil. These theorems wili be arranged in 
groups, according to the method of proof to be used, 
followed by a group of general ur mixed exercises. 

Le.t the pupil form a list of the, conditions that make 
two triangles equal. 

(See Arts. 96, 97, 1)8, 101, 102, 141, 142.) 

£XERCIS£S. CROUP 4 

EQUALITY OV 'rilIA.\'Gl.ES 

Ex, 1. Riven AHC any triangle, BO Iha hi- £ 

seotor o( I ABC, and JD 1 BO; proved J B0== 
A BOD. 

[SUG. In the A AffO and nno wbal Hues 
are equal! What A are equal t etc.] 

Ex. 2. If, at any point in the biseolov of 
an^ie, a X be ereeted and produced to meet t 
Bides of the angle, how many triangles bvo forint 
Are these triangles equal ? Prove this. 



Ex. 3. If, through the midpoint of a f 
straight line, another line be drawn, and prnduc 
to meet the perpendiculars erected at the o( 
of the given line, the triangles so formed i 



P^ 



sect each other and their 
of equal triangles are 



Ex. 4. H two straight 1. 
ties be joined, how many 
Prove this. 



Ei. 5. It equal segments from the base be 
of an isosceles triangle, and lines be drawn from 
segments to the opposite vertices, prove that t 
angles are formed. 



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ESEECISES. EQUALITY OP TRIANGLES 85 

Ex, 6. If, upon the sides of an angle, equal segmnnts be laid off 
from the vortex, and lines be drawn fram the ends of these Begments 
to any point in the bisector of the angle, prove that the trianglea 
formed are equal. 

Ex. 7. If two sides of a triangle be produced, each its own 
length, through the vertex in whioh they meet, and the extremitiea 
of the produced parts be joined, prove that a new triangle is formed 
whieli equals the original triangle. 



Ex. 8. Given AB^DC, and BC^DA; yro- 
ABAC=A1>AC. W 
triangles is tbare in the figure 



, and BC^ DA : prove /\g-^ 



Ex. 9. Two right triangles are equal if ilieir corresponding legs 



Ex. 10. The ttltitudsH from the extremities of the base of an 
isosceles triangle upon the legs of tlie triangle divide the tiguve into 
how many pairs of equal triangles ! Prove this. 

Ex. 11. In a given quadrilateral two adjacent aides are equal and a 
diagonal biseots the aogln between tlie^e .■sides. Prove tliat the diagounl 
bisects the qnadrilaternl. 

Ex. 12. If, from the ends of the shorter base of an is os eel es trape- 
zoid, lines he drawn parallel to the legs and produced to meet the 
other base, prove that a pair of equal triangles is formed. 



189. Prooi of the equality of lines. Thei-c are several 
methods of proving tiint two lines (segments) are equa!. 
One of the principal tnetJiods of proving that two lines are 
equal is J>y provitifj that two triangles, in tvMch the given 
lines form homoloijoit» /uirls, are equal. 



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EXERCISES. CEtOUP a 

EQUATiiTY OK j.ixr:.-; 

Ex. 1. Given ABC niiy t.-ismsle, JIO the h)- 
BPPtorof lABC, and -J/) ± HO; prove AO = On. 

Ex, 2. K, at any point in tbe bisector of so 
angle, a porpendicular be ereeted to the bisector 
and produced to meet the sides of tho angle, 
the perpendicular 13 divided into two equal parts at 

Ex. 3. If two aides of a triangle be produced, 
each its own length, tbronyh the common vertex, 
the line joining tbe extremities of the produced 
parts equals tbe third side of tbe triangle (i e., 
!ȣ = BA). 

Ex. 4. It eqnal segments from the base bi- liijil 
otT on llie legs of an iaoectles triangle, lined dr;i«n 
these segments to the opposite vertices are isqual. 

Ex. 5. Given AR\\ QB, and AF = FJ1: 

prove BF = PQ. 

Ex. 6. The bisector of the vertical an- 

gleof an isosceles triangle bisects the base. '^ ■" 

Ex. 7. Tiie aititiides of an isosceles triangle upon the 



eqnr 



rectangle are equal, 
isosceles triangle to tho legs 
nt a triangle are equal, the 1 



Ex. 8. The diagonals of 

Ex. 9. The medians of i 

Ex. 10. If two altitude- 
ioaeeles, 

Ejt. 11. Tbe perpendiculars to a diagonal 
f a parallelogram from a pair of opposite 
ertiees are equal. 

Ex. 12. If the equal sides of an isosceles triangle be prod 
hrough the vertex so that the produced parts are equal, the lines 
□g the extremities of the produced parts U> the extremities o 
lase are equal. 

Ex. 13. If tbe base of au isosceles triangle be trisected, 
IrawL from tho vertex to tbe points uf ti'iaectiun are equal. 



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EXEECISES. EQUALITY OF ANGLES 



87 




Lines may also be proved equal by showing that they are: 

opposite equal angles in a triangle; 
or opposite sides of a parallelogram; 
at parallel lines comprehended hetweeit parallel lines, etc. 
(See Arts. 100, 155, 157, ete.). 

Ei. 14. If the exterior angles at the base 
of a triangle are equal, the trian- 

Ex. 15. In the biseotora of 
the equal angles of an isuseftles 
triangle, the segments nest to 
the base are equal {AO= OP). 

Ex 16. In A ABC, given 

AJ) = AC. DE II BC; prove 
Ah = AE. 

Ex. 17. Given AB = DC, 
a.)iABC = AD; provf, AE= EC. 

190. Proof of the equality of angles may be obtained 
in several different ways. 

One of the principal methods of proving that two angles 
are equal is by proving that two triangles, in tvMch the gieen 
angles form Jyyimlogous parts, are equal. 



EXERCIGES. CROUP' & 

EQUALITY OF ANGLES 

i1. 1- Given J^r: any A, BO the bisector 
of the lABC, amlJDX/iO; prove lliAO 
= ^ BDO. 

Es. 2. If, at any point in the bisector of an 
angle, a perpendieular bu erected and product;J 
to me^t the sidea of the anjtle, the ptr- 
pendieular makes equal angles with the sidett 
of the angle. 

Ex.3, Given AH-liV, M^ AD - BV; 
pruvb IM~/ n. 



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88 IJOOK I. I'LANE GEOJILTJIV 

Ex. 4. If equal segments from the base be laid olT on Iho li%'3 of 
an isoswlea ti'iaiigle, the lines <lrawn from the extreniilips of ibe 
segments to the opposite vevtii;es miike equal anglus vvilh llie base. 



Ex. 6. The altitudes upo 
equal angles with tbts base. 



Ex. 7, Theai.tgoiia 



Aiij^los may iilso he proved cqivA 



aft- opjMsite equal sides in an i^oKcrli-i^ li-ianyh; 
or uir veriical angles; 

or ore complements [or supfiUmenlx) of equal ani/les; 
or hij the use. of the propeHies of parallel lines; 
or that their sides are parallel, or perpendicular. 



(Sco Arts. 



78, 99, 124, 1120, 130, 132.; 




I isosBfiles ti-ia,ngle the txteiior auf^li 
■ing the base are equal. 

E*. 9. Given AC = CB, an.t DE 
Jill; prove Z.CDE = ZCJSn. 



',x. 11. Conversely given 
= Z B, and CE \\ AB ; 
■e that CE bisects I DCA 



Ex. 12. Given BD tbe bisector of thi 
angle ABC, and PR \\ CB; prove PJIB 
isoseelea A. Let the pupil state ibis theuri 
in general language. 




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EXEKCISES. PAEALLEL LINEH 




hig ihe a'ier- 



Ex. 13. Given AB = AC, 
aii3BD=CE; prove ZBCfl= 
Z CBE. How many pairs of 
equal A in the figure f of 
equal lines? of equal A1 

Ex. 14. A line drawn 
through the vertex of an angle, 
perpendicular to the bisector 
of the angle, makes equal 
angles wilh the sides of the given angle. 

Ex. 15. If a Btraigbt line which biaecis 
one of two vertical angles be pi'oduued, it 
biseots the other vei'tioal angle also. 

191. Proof that two lines are parallel i 
bj' showing that: 

itie Ihu'fi lire cul ha a l7-(ntsi'ernfil, » 
nate interior angles equal; 
or making the extenor interior angles equal; 
or making the interior angles on the same sitle of the 

transversal supple^iieniary ; 
03- that the lines are opposite sides of a parallelogram; 
or tJuit one of the lines joins the midpoints of tico sides 
of a triangle, and the other line is the third side of 
the triangle. 
(See Arts. 125, 127, 129, 147, 177, ett.} 



EXERCISES. CROUP 7 

PARALLEL LINES 
Ex. 1. If two sideg of a triangle be produci'ii, each 
vertex, the line joining their ex- 
tremities is parallel to the Ihird 
side of the triangle. 

Ex, 2. The bisectors of two 
alternate interior augles of pui- 
allel lines are parallel. 
A = ZS, and IDCE = lECA ; 



-Ex. 3. Given 
rove CE \\ BA. 




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\n\ HOOK 1. PLANK iu:o.Mi/n;v 

Ek. 4. The hi life tors o£ the opiiositci aiiflea oC a ]Kir;ilkloi;ram are 
Ex. 5. l.im.a perpcu.li^iilar lo parallel Imp-. Krs pamllel (or 
B paiiillfl if Iwo points on oua iiue 

192. Tlie proof of a numerical property of the recti- 
linear figures (of Book I) usually depends on one of the 
following; 

The sum of the angles abmit a given point on the sami- 
side of a straighl line passing through the point is 180 ; 
or the smn of the angles about a point is 3G0°; 
<ii' the sum of the angles of a triangle is 180° ; 
or the .iiiiu of the interior angles of parallel lines on the 

same side of a transversal is 130°; 
or file sum of the interior angles of a polygon of n side^: 

is (ft--2) 180°; 
:;i- the Slim of the erlertor angles of a polygon is 360°. 
(See Arts. 76, 77, 128, 134, 172, 175.) 

EXERCISES. CROUP 8 

^;L■.■^[EKK'AI. rROl'KKTlKS 



Ex, 1. It an e-ttfirior angle o£ a triaiitcle is "123° and an opp^ 
inferior angle is SK", find the other two angles of the triangle. 

Ea. 2. Find the angle formed by the bis'eotoi-B of the two s 
bugles ot a right triangle, 

Ex.3. If two angles of a triangle are 50" and 60", find the s 
formed by their biaaotora. Find the same if the two angles toi 

Ex.4. It -he vertm aiigle of an irascejes triangle is 4(1" a 
;pei'pendieiilar I. drawn from an extremity o£ the base to the opp 
■liile, fliiU Ibe ai.^^lea of the figure. 



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EXERCISES. NUMEKICAL PROPERTIES 



Ex. S. 


If til.- v.rfti'X aiife-le of an isosceles ti-iangl.) 


^40" 


find tl 


^le me 


tidfd between the altitudes drawn from tlie 
tiie oppoiiite sitles. 


Bxtve 


mitiea o 


Ex.6. 


How many degreaa in eac:h angle of an equiu 


ij:;ula 


 doiieea 


n? of 


n equiangular li-gon? 






Ex.7 


How many diagonals are there in a pcntaf-o 
decagonT inanx-gon! 


uT i 


a hexn 



The methods oi proving that a given angle is a right 
angle (or that a given line is perpendicular to another given 
line), or that one angle is the supplement of another, -aiv 
closely related to tlie ubore tHethuiU of ubtiiinhuj llw iiiiiHi-ri' 
(■(tl v(il„.''8 of (jivi'H iiiiglef<. 

Ex. 8. .\i.y iniir of adja.'.^iit aiigleH of ;. parallelogram is mipplu- 



Ex.9. If oneaugleof aparalie 
a rectangle. 

Ex. 10. The bisectors of two (, 
adjacent angles form a riplit angle 


ograin is a rigli 

upplementary 
(ave verpOii- 

iterior angles o 


angle the fignre is 


Ex. 11. The biseetors of two i 


the same side of a 



Z.^ form a right angle. 
Ex. 12. If one of the It'gs 

(JB) of an isosceles triangle 
he produced its own length {ISD) and its ex  
tremity (D) bo joined to the other end of the base 
(C), the line last drawn {DC) is perpendicnlar to 
the base. 

193. Algebraic method of proving theorems. The proof 
ci certain properties of a geometric ligiu'ti is ut'teii fin-ilitiitfil 
ty the use of an tilgehraic si/iiihoJ for un Kiil.iioini aiKjh "i- 
tm itjik'iiown line of the. fiifKir. (i>id tin' iifit of uit i<jutilion or 
ether ulyebrak metkud t/J solution. 



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;)li I'.OOK I. I'LANE (iEUMlITKi' 

EXERCISES. CftOUP 3 

Al,i;EHtiAIO Mirnillll 

Ex. 1. Find the number o[ degrees in :iii an^le wlii'.'h eqiiiils twice 
its complement! 

[Srti, Let ^ = 1116 complement, etc.] 

Ex; 2. Fiud the number of degrees in nn nugle which equals its 
supplement? in one whk'h equals oue-third its supplemeut! 

Ex. 3. The angular space Kbout a point is divided into four angles 
TObieli are in the ratio 1, 2, 3, 4. Piud the number of degrees iu eaeU 
angle. 

[Sl-o. j:}- 2.i+:!i--t-4J^ = 3tiO°, etc.]. 

Ex, 4. The ougles of a triangle are in the ratio 1 , 2, a ; lind the 
angles, 

Ex. 5- Two angles are supplementary and the greater exceeds the 
less b>- ^O" ; find tlie angles. 

Ex, 6, Find all llie .angles of a paraUelogram it one of tiiem is 
double iLiiother angle. 

Ex. 7. One of the base angles of a triangle is double the other, and 
the exterior angle at the vortex is 105°. Find the angles of the triangle. 
Ex. 8, How many sides has a polygon the aum of whose angles is 
fourteen right angles? 

[Sl-g. 2(h-2) = U; fimln.] 

Ex. 9. How many sides has a polygon the sura of whose angles is 
ten riglit angles! twenty right angles! 720°i 

Ex. 10. How many sides has an equiangular polygon one of whose 
angles is seven -fourths of a right angle f 

Ex. 11. How many aides has a polygon the sum of whose interior 
angles equals the sum of the esterior angles ! 

Ex. 12. How many sides has a polygon the sum of whose interior 
angles equals three times the aum of the exterior angles? 

Ex, 13. If the base of any triangle be produced in both directions, 
the sum of the exterior angles thus formed, diminished 
by the vertex angle, is equal to twi right angles. 
[Sua. 180°— a-i-lSO'^— ft-tlSO"— a— (i)=,6te.] 



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EXERCISES. AUXILIARY LINES 



Ea. 14. TliO bisectors of the base angles of a 
iaoscelea triangle include an angle -which is equii! 
the exterior angle at the base. 

[Sm. To prove « = '!>, denote one of the baso A. 
2x, etc.] 

Ex.15 In an i-osccles triangle the altitude upon 
one of the legs makes an angle with the base which 
eqaala one-half the lertcx angle. 

[Suu To proie «^ Wi, show that n = 90°— r, i = lSO° 

Ex. 16. If the opposite angles of a qiiadri 
lateral are eqaal, the figrure is a, parallelogram 




194. Use of auxiliary lines. laequalities. The demon- 
stration of a property of a geometrical figure is frequently 
facilitated by drawing one or more auxiliary lines on tlie 
figure. For examples of the nse of such lines, see Props. 
Ill, V, IX, etc., of Book I. 

Some of tiie principal auxiliary lines ii!>eil on reutiUucar 
figures are: 

a line connecting iwo given points; 

a line through a given point parallel io a (liven Vin< ; 

a line- through a given point perpendicular to a gicen line; 

a line making a given angle with a given line; 

a line produced its own length, etc. 



EXERCISES. CROUP 

AfXILlARY LINES 



2i£. 1. In the quadriiaterul 

AB^AD, and 110 = 0); prove 

[SuG. Draw AC, etc.] 

Ex, 2. Prove that the nn(;h 

of an isosceles trupezoid art S' 



IJI = / II. 




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04 BOtHi I, I'L.VKK (IKUilETKV 

Ek. 3. titiile and prove the converse of Ex. '1. 

^ El. 4. GivRii J JS jl ClJ ; provR z ?i = Z ,< 

^-■>,,^^ Ex. 5. Conversely, Kiveti Z ?^ 

'^ — "■'^ = Z ,1 + Z c ; prove JH |1 CD. 

Ex. 6. Tlio median to the hypotenuae at a ri;;!:! 
tiiaiiglti is oae-hal£ the hypotenuse. 

Ex. 7. If one acute angle of a right triangle is double 
the other, the hypotenuse is double the shorter leg. 
[Suu. Draw the median to the hypotenuse, etc.] 



Ex. 8. In an isoBeeles triangle, 

the Slim of the perpendiculars 

drnwii from any point in the base 
I to the legs is equal to the altitude upon one of 
U the legs. 




la some cases it is useful to <iraiv firo < 



Ex. 9. Show that the median of a trapezoid equals one-half the 
siun of Ihe two bases by drawing a line through the miiipoirt of one 
IfC of the trapezoid, parallel to the other leg and meeting one base 
aud the other base produced. 

Ex. 10. Lines joining the midpoints of the sides of a quiidrilatei-ni 
taken in order form a parallelogram. 

[Sui!. Draw the diagonals of the quadrilat- 
eral and use Art. 177.J 



Ex. 11, If the opposlti 
sides of a hexagon are equa 
and one pair of sides (.4B ami jj; 
CD) are parallel, the opposite anglcB of the he.-tagoi 

Ex. 12. 'Hteu Allelic, and An=CJi; prov 




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EXERCISES. INEQUALITIES Vii 

Let the student form a list of tlie principles proved in 
Book I concerning unequal lines ottd unequal angles. 
(See Arts. 92, D3. 95, etc.). 

EXERCISES. CROUP 19 

INEQUALITIES 

Ex, 1. In the triangle ABC let i' be any pnint in the Bide BG; 
prove All + BC > AP '.- PC. 

Ex. 2. In the quadrilateral Alscn let F be any point in the side 
.Wf; prove perimeter of APCD > perimeter o! AfD. 

Ex 3. In the triangle AISC let 1) he any point in AB nnd F any 
point in IW. Ftove AB + BC > An + I)l'-+ FC. 

Ex. 4. The sura of the four sides of a (piadri lateral is g.eattr IIihu 
the Bum of the (ilagonala. 

Ei, B. If, from any point within a triangle, liiiea be diaivn to the 
verfiees, the sum of the hues drawn is greater tbau one-half the sum 
of the sides of the triangle. 

[SUQ. Use Art. 92 three times, etc.] 

Ek. 6. If, from any point, within a triangle, lines be drawn to the 
vertices, the sum of the lines drawn is less than the Bum of the sides 
of the triangle. 

[Sue. Vse Art, 05.] 

Ex. 7. The median to any side of a triangle 
leas than half the sum o£ the other two sides. 

[SiTG. Prodnee the median its own length.] j 

Ex, 8. If B is the vertex of an isosei'les triangle vlBr, r. 
produced to the point J), lliAti ia greater than IBIIA. 

Ex. 9 lu the figure 51. 7],7t(.' > AB; prove llitV yr* 
/.BVA. [Sifi, Use Art. 108 } 

Ex. 10 111 the same figure, show that /■'(' < /.'' . 
Ex. 11. Ill the qimilrilalet-al 
liV the shortest : prove ,'- .(/((-srej 
llian I BAD. 

Ex. 12 Lines are drawn from .i, ff, ('. ?i, Coiirpoinls in a 
line, to the point f ontside of the line Wlucli anj;lo3 on 1 
atu lefiS than angle ACF^ Which angles are greater I 



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1 '1 1 BOOK T. I'T.AXE GP:OMETI{Y 

195. Indirect demonstrations. Loci. Tii Book I tliri'e 
nnjthnds of indirect proof hiive been used. 

1. The reduction to an absurdity {rednctio ad absiiv- 
dum), that is, the proof that the negative of a given theorem 
leads to mi ubsuvdity Csee Prop. 5). 

2. The method of exclusion, that is, showing that any 
other statement ihtn the. i/iven theorem cannot he true (see 
Pi-ops. XV, SVII). This method is a special ease of the 
preceding, the negative of a given theorem being divided 
in it into ti^'o parts ■which are separately shown to be 
impossible. 

3. The method of coincidence, that is, proof that a given 
line coincides idth annthrr line, irlii'-Ji fulfils cvrioin re- 
quired cmiditions (see Props. XVII, XXIIJ, XXV, etc.). 

EXERCISES. CROUr J3 

INDIRECT, OE NEGATIVE DEJIONSTKATIONS 
Prave the folloiyiiig by iiu uvlitneX method: 

Ez. 1. Every point within an angle and not in the liiseetor of the 
angle is unequally distant from the sides of the an^ic. 

[Si,"Q. In the given angle take P any point not in llie bisector of 
the angle. Then, if P is not unequally distant from AG and on, it 
must be equally distant from them, etc.] 

Ex. 2. I£ two straight lines are cut by a transversal, maldug the 
al:ernate interior angles nnequal, tie lines are not parallel. 

Ex. 3. The line joining the midpoints of two sides of !i triangln 
is parallel to the thii'd side. 

[Sua. Through one of the midpoints draw a line |1 to the Ihini 
side, show that it bisects the seeond side and that the line joiulug the 
midpoints eoinaides with it. J 



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EXERCISES. LOCI 



97 



Ex. 4. IE, ffom a point P in a line AB, Jines PC and FD be drawn 
on opposite sides of AB making the angle AFC equal to tiie angle BPD, 
PC and PD are in the same Htraiglit line. 

[SuG. From P draw PQ in the same straight line with PC and 
show that PD eoineideB with it,] 

Ex. 5. The hlaectors of two vertical angles are in the same 
straight line. 

Ex. 6. In the triangle ABC, Bis any point in the side AB, and Eis 



any poi 



Ex. : 
a given 



II tho side 



Proi 



i that BE and IJC cannot bisect each 



Whnt 



 Vi-< 



EXERCISES. CROUP 1-3 

LOCI 

he iiifus of iill points :d a pivf 
e this. 



Ex. 2. What is the locus of all points equidistant from two given 
parallel lines f Prove this. 

By use of known loci (see Arts. 112-118), prove the following: 

Ex. 3. The diagonals of a rhombus are perpendicular to each 
other. [Su«. See Art. 113.] 

Ex. 4. The median of tin isosceles triangle is perpendicular to the 

Ex. 5. The line that joins the vertices of two isosceles triangles on 
the same biiRe is perpendicular to the base. 

1 96. General method of obtaming a demonstration of a 
theorem. Analysis. 

A due to the sointion of some of the raoi-e difficult 
theorems is often obtained by proeeeding thus: 

Assume the proposed theorem as true; observe tchai other 
felaiion among the parts of the figure must then be true; 
proceed backit-ard thus, step by step, till the required theorem 
1^ found i-o depcwl on some linoirn truth; then, starting with 
this knoivn truth, reverse the ste/is tiilrn, (iiid thus build up 
a direct proof of the nqnireil theorem . 

This method is called solution by analysis. 



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98 BOOK 1. PLAXE GEOMETRY 

The following is a simple example of (lie 
method : 

Ex. Given AB and AC Hie logs of an ienacoloa 
triangle and D any point on AD; prove DC greater 
than DB. 

Analysis. If DC > DB, 

I Bis greater than I DCB. Art, 104. 

Hence subatituting for ZB its equal, lACli, we 

have Z^C/; is greater than ^DCB, Ax. 8. 

But we know that lACB > IDCB. As, 7. 

Hence, Direct Proof (or Synthesis) 

ZACB is greater than ITlCn, Ai. 7. 

.-. Z B is greater than IBCE. Ax.. 8. 

:.DC > DB. Art. ICU, 

0- E. D. 

The first part (aualysis) of the above process is to be 
purely mental work ou the part of the pupil, in investiga- 
ting a given theorem; the second part (the direct proof, or 
synthesis) is to be written out as the required solution. 

In working the following exercises, this method will be 
found to be necessary in the solution of only a few of the 
more difficult theorems. 



EXERCISES. QFiOUP 14 
THEOREMS PROVED BY VAKIOUS METHODS 



Ex. 1. If two opposite 
diagonal connecting thei 
tills diagonal. 



;a of a quadrilateral are blaected by the 
ices, the quadrilateral U bisected by 



Ex. 2 Perpendipuiars dras 
triangle to the median to the b 



a the extremities uf the base of a 

Ex. 3. If the perpendiculars from the extremities of the base of a 
triangle to the other two sides are equal. (1) these perpendiculars make 
«qual angles with the base, (a) the triangle is isosceles. 



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MISCELLANEODS EXEItCIBEH 
Ex.4. I( tlie lines Jfi and CD iulorse 



rsect.tbon Ali+Cn > AC'+DB. 

iviAes trLacgle is 44^, and one ot 
praiiuced to me el the opposite 



Ek, 5. The verlPK aii)|l« cf an u 
the base anplea is bisected by a hn 
side. Find ad the angles of the figu 



Kx,6. In the figure of Prop, V prove that ZJPC is greater than 
I ABC. Also prove the same m another wiiy by means of an auxiliary 
line drawn througli fi and F, 

Ex. 7. Ptrpeudioiilnrs drown from the mi.lpoinl of the ba.^ie of a:a 
iaosoeles triangle to the leRs are etinal. 

Ex. 8 State and prove the 

Ex. 9. In an Isosceles triat 



Ex 10. In a re ent 

angle at the re entrant 



t quadrilateral the exterior 
tes equals the sum of the 



dueedtoD; prove BD > JD > JB, 

Ex, 12. If irOLn a point in the liisector ul au obliqiic aiiglu 
formed is a rborabus. 



Ex 14. It the median nf a triang 
the triangle is isosceles, 

Ex. 15. Given JB = Ja 
lBAC = ilB, 
and DFX ISC, 
prove A EFA equilaleral. 
, Unalvsis. 1{ a fJA-i>equilatei 



= dO^ 



Her 



"sing iLbAC= ilC.\ 



% din 




U, lfEA = W°; .■.Z»Kfi = GO°; 
proof, show that IC ~ 30° by 



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1 00 BOOK I . I'iANE L"; KOMKTliY 

Ex. 16. If Ihe ilingooals o£ a quadrjlatoral bisect each other at 
right angles, what kiad of a figure is tlie quaJrilateval f Prove this. 

Ex. 17. From the point in which the altitudes drawn to the legs ot 
an isosceles triauf;le intersect, a line ia drawn to the vertex, Prove 
tiat this lino bisects the attgle at the vertes. 

Ex. 18. If from a point within an acuta angle perpendiculars ai'e 
drawn to the sides o£ tlie angle, the angle formoii by thesa perpendictt- 
tara is the supplement of the given angle. 

Ex. 19. Linos joining tlio midpoints of tli 
aides of a triangle divide the triangle into Ecu. 
eqaa! triangles. 

Ex. 20. If, in the parallelogram ABCD, 
BP = VQ, then AQCF is a parallelogram. 

[A>-.\LYsiS. If J^CPisa^:?, JP=find 
II QC. .'. hegin the direct proof by sliow 
ingthat JP=andi3 || QC] 

Ex. 21. If the diagonals of a parallelogriim are equal, what kind 
of a figure is the parallelogram! Prove this. 

Ex. 22. If the cngle A of the triangle AUG is 50° and the oxterior 
angle BCD is 1L'0°, which is the largest side in the triangle ! 

Ex. 23. Two triangles are eqnal if two sides and the median to 
one of these sides in one triangle are equal, respeistively, to two 
homologous sides and a median iii the other. 

Ex. 24. Two isosceles triangles are eqnal if the base and an angle 
of one are equal to the base and the homologous angle of the otiier. 

Ex. 25. Two equilateral triangles are equal if an altitude of one 
IB equal to an altitude of the other. 

Ex. 26. If two medians of a triangle are eqnal, the triangle ia 
isosceles. 

[SuG. On Fig, to Prop. XL VI, taking AD= EC, prove A AOC isos- 
celes, AAEC=l\dDC, etc, How could this theorem be investigated 




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Miscellaneous exekcises 



Ex. 27. Prove the sum of the aiif-lea of a 1 
right angles by drawing a liue through the ve 
parallel to the baae. 


:ri:mglft equal to tw 
rtes of the ti'iangl 


Ex. 28. The homologous mediaus o£ two 
equal. 


equal triangUa ar 


Ex. 29. The bifleetora of nn angle of a tri- 
angle and of the twn erterloi- aiigios at the other 


A 


Ex. 30. If JKC is an equilateral triangle nTul 
AP= BQ= CE, then I'QIt is an equilateral 




triangle. 





Ex.31. If the two baae ancles of ft trianglo 
l)e bisected, and through the point of ii;tei'See- 
tion of the two bisectors a line be drawn parallel 
to the base, the part of this line intercepted 
between, the two sides equals the sum of the 
aegments of the sides included between the par- 
allel and the base (i, e., prove PQ=AP+ QC). 

Ex. 32. Two quadrilaterals are equal, if three sides and the two 
included angles of one are equal to three aidea and the two included 
anglea of the other, respective!}'. 

Ex. 33. If the diagonals of a quadrilateral bisect each other, the 
figure is a parallelogram. 

Ex. 34. Lines joining the midpoints of tho sides of a Vtctangle 
in order form a rhombus, 

Ex. 35. Lines joiuing the midpoints of the sides of a rhombus 
l^orm a rectangle. 

Ex. 36, The bisectors of th? angles of a paraliclogram form a 
reetangle. 

Ex. 37. The bisectors of the angles of a rectangle funn a square. 

Ex. 88. If lines be drawn through the v 
parallel to tho diagonals, a parallelogram ii 
M large as the original quadrilateral. 



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1>I,ANE GHOMK'I'RV 



grata to thtt luuipo'iit'i of a, pair of opposite sides trisect; 
of the pavallelogram . 

Ex. 40. Oq the liiaKOnal AC ot a paralli4osram AltCD ei) 
AP and CQ, are marked oil, Prove Bl'HQ a jwi'allelogl'a 
many pairs of equal friaiiglps does the ligiive i-oiitaiii i 



Ex. 41. Tlie oppo! 



Ex, 42. In an isosed™ trapezoid, tiie diagonals are equal. 

Ex. 43. If the upper base of an isosceles trapezoid equals the eum 
of the legs, and lines be dran-n from the midpoint of the upper base to 
the exiremities of the lower bas^e, how mauj' iKoseeles triangles are 
formed ! Prove this. 

Ex. 44. The lines joining the midpoints of the sides of an isos- 
celes trapezoid, taken in order, form a rhombus ov a squary. 

Ex. 45. The bisectors of the angles of a trapezoid form a quadri- 
lateral whose opposite angles are supplementaif. 



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Book U 

THE CIRCLE 

197. A circle is a portion of a plane bounded by a 
curved line, all points of wliiisli are equally distant from a 
point within <mllod the center. 

The circumference of a (linUe is the curved line bounding 
the circle. The term itin^le may also be used for the 
bounding line, if no ambiguity results. 

A eii'ple is named by nomiiii; its center, an the eircile O; or by 
naming two or more points on its cireumfereuce, ag tlio uirtlo AVD. 



198. A radius o£ a circle is a 
straight line drawn from the center 
to any point on the circumference, 
as AO. A diameter of a circle is a 
straight line drawn through the cen- 
ter and terminated by the circumfe- 
rence, as BC. 

199. An arc is any portion of 
a circumference, as AG. A semi- 
circumference is an arc equal to 
one-half the eireumfei-ence, as 
BAG. A quadrant is an arc equal 
to one-fourth of a eireumferenee, 
B.SBD. 

200. A chord is a straight line 
Joining the extremities of an arc, 
as EF. 




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104 BOOK Tr. PLASE (lEOMETKV 

Every chord subtends two arcs. A miuor arc is the 
smaller of two ares subtended by a chord. A major arc is 
the larger of two arcs subtended by a chord. Thus, for the 
chord EF the minor arc is EPF, and the majnr arc is ETF. 

Conjugate arcs is a general term for a pair of minor and 
major aros. 

If the are subtended by a given chord is mentioned, 
unless it is otherwise specified, the minor arc is meant. 

201. A tangent to a circle is a straight line which, if 
produced, has but one point in eonunon with the circle, aa 
MS. Hence, a tangent touches the circumference in one 
point only. 

A secant is a straight line which, if produced, intersects 
the circumference in two points, 'as GH. 




yis. 3 Fis. * 

202, A segment of a circle is a portion of the circle 
bounded by an arc and its chord, as ABO (Fig. 3). 

Into how many segments does each chord divide a circle? 
A semicircle is a segment bounded by a semicircumfer- 
enee and its diameter. 

203. A sector of a circle is a portion of a circle bounded 
by two radii and the are included by them, aa FOQ 
(Fis, 3), 



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THE cmcLE l05 

204. A central angle is an an^lo whose vertex is at the 
center and wliose sides are radii, as the angle POQ 
(Fig. 3). 

An inscribed angle is an angle whose vertex is in the 
circumference and whose sides are chords, as the angle A/JG' 
(Fig. 4). 

An angle inscribed in a segment is an angle whose ver- 
tex is in the are of the segment and whose sides are chords 
drawn ft'oni the vertex to the extremities of tlie are. Let 
the pupil draw a circle, a segment io it, and an angle In- 
scribed in tlie isi'gnieut. 

205. Two circles tangent to each other are circles which 
are tangent to the same straiglit line at the same point. 
They are tangent hifernally or externally according as one 
circle lies entirely within or entirely without the other. 
See the figures, page 122. 

Concentric circles are circles which have the same center. 
Let tlie' pupil draw a pair of concentric circles. 

206. A polygon inscribed in a circle is a polygon all of 
whose vertices lie in the circumference of the circle, as 
ABODE (Fig. 4). 

A circle circumscribed about a polygon is a cii'ele whose 
circumference passes through every vertex of the polygon. 

207. A polygon circumscribed abont a circle is a polygon 
all of wliose sides are tangent to the circle, as PQlttiT 
(Fig. 5). 

A circle inscribed in a polygon is a circle to which all 
the sides of the polygon are tangent. 

Coacyclic points arc poiuts lying on the same circum- 
ference. 



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1U6 BOOK n. I'LAKE GEOMETRV 

PROPERTIES OF THE CIRCLE INFEERED IMMEDIATEIY 

208. liiidii of lli<: same circle, or of equal circles, are 
equal. 

209. The flianicier of a circle equals fu-ice its radius. 

210. DhiDiefera of the same circle, or of equal circles, 
are equal. 

211. If two circles arc equal (i. e., ma}/ be made lo 
coincide. Art. 13), their radii are equal, and conversely. 

212. A diameter of a circle bisects the circle. For, by 
placing the two parts of the circle so that the diameters 
coincide and their arcs fait on the same side of the diame- 
ter, these arcs will coincide (Art. 197). 

213. A straight line cannot intersect a circle in more 
than tivo points. For, if a straight Hue can intersect a 
circle in three (or more) points, three or more equal lines 
(radii) can be drawn from the same point (the center) to 
the straight line. But this is impossible {Art. 111). 

Proposition I. Theorem 

214. -4 diameter of a circle is longer than any other 

chord. 




Given AB a diameter, and CD any other chord in the 
circle 0. 

To prove AB > CD. 



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THE CIRCLE 



107 



Proof. Draw Uit radii OC and O'D. 
Then, in the A OCD, OG+OD> CD. (Whyf) 

Substituting for OC its equal OA, and for 01} its oqud 
OB. Ax. s. 

0A + Oh, or Ali> CD, 

Q. E. B, 



ri!Oi'OfilTIOX If. Tbeoukm 

215. 7/i tiie same circle, or in equal circles, equal cen- 
tral amjhs i)i/crcepl cijuul arcs oil the circumference. 





Given the equal eircios and 0', aud tlie equal central 
A. AOBaviAA'O'B'. 

To prove the are AB=are A'B' . 

Proof. Apply the circle 0' to the circle so that the 
center 0' coincides with the center 0, and the radius O'A' 
?.'ith the radius OA. 

Then the radius O'B' will fail on the radius OB, 

iforhjhi/p. lAOB^lA'O'B'j. 
And B' will fall on B, Art. 20fi. 

(/or OB = 0'B', being ra<Ui o/i-<pi<i! ®). 
Hence arc A'B' will coincide with AB, Art. I!i7. 

(/or uU poiiils of each arc arc iijiiUUslanI from (JiC i-'cnffr). 

.'. are A'B'= arc AB. Art. 4T. 

Q. £. D. 



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108 l;OOK n. I'LANE UEOME'IliV 

PrOTOSITION III. ThEOEEM (CONVEE5G OF pROP II) 

216. In the same drch, or in eguai circles, equal crn-n 
subtend equal augks at the center. 





Given the equal (;irek's niul 0', niiil the efjual ares 
AB and A'B' subteiuliug the central A ami 0'. 

To prove ZO = ZO'. 

Proof, Apply the eirele 0' to the equal circle so that 
the center & colncii^es with the center and the point A' 
with the point A. 

Then the point B' will fall on B, 

(for arc A'B'^arc AB iy hyp.). 

Hence the radius ffA' will coincide with OA, and radius 
O'B' with OB, Avi. CA. 

{betuieeii two points only one straight line can he liAiini). 
.'. L and £ 0' coincide. 

.-. 10= 10'. Art. 47. 

0, E. D. 

217- Cor. In the fame circle, or in equal circles, of 
two unequal central angles the greater angle intercepts the 
greater arc, and, conversely, of two unequal ares, the greater 
arc subtends the greater angle at the center. 

Bx, Draw a, circle and hi it a segment whicb la less tbau tlie 
sector hiviug the same arc. Also oue tiiat is greater. A\ea one that 



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THE CIliCLE 109 

Proposition IV. Theorem 

218. In fhe snme. circle, or hi eqmil circles, equal chords 
subtend equal ai-cs. 





Given the equal eirelos; and 0', uud the elioi'd AB = 
chord A'B'. 

To prove are AB = arc A'B'. 

Proof. Draw the radii OA, OB, O'A', O'B'. 

Theu, in the A AOB and A'O'B', 

AB^A'B'. (Wbr I) 

AO=A'0', and BO=^!i'0'. (Why?) 

.-. A AOB^AA'O'li'. (Why?) 

.-. Z0= 10'. (Why J) 



e O, . 



;, arc AB = ai-c A'B'. Ari 

y fv/iio; ®, eqiKil cmibut A intercept equal a 



215. 



Ex. 1. I„ the above f.^'ure, if thord AB=l iu., chord . 


XB'--=. 


aDaare^B = Uin., find the lenRth of iire .J'JJ'. 




Ex. 2, Draw a clrtlt aud mark off a part of it that is 


both V. 


ment HDd a seolor. 




El. 3. If the diatance fvoiu the eentor of a circle t 


.0 a li 


greater than the radius, will tha line iiitetWHct tUo ciri:HHi 


fercuui 



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10 UOOK II. PLAMi GEOMETllY 

Piiopoarnux V. TheoiU':m 

219, III ill'' s"»f'' rlrrlr, '/r in equal circles, equal arcs 
>■€ suhlcnded h>j equal eJiurds. 





Given the. equal circles O and (T, amlara AB^iiva A' B'. 

To prove oliord AD — eliord .1'/!'. 

Proof. Draw the radii AO, BO, A'O', B'O'. 

Tlien, ill tlie A AOB and A'O'B', 

10 = Iff, Art. 216. 

(for are AJi = A' B' , iiiid, in Ilia snmo O, or in equal ®, equal a 



U-ml equal A at ihe center). 






Also OA = 0'A', and OB^O'B'. 




(Why ?) 


.-. A AOB- A A'O'B'. 




(Why ?) 


.-. AB^A'B'. 




(Why f] 




A'J! 


Q. E. B. 


Ex. 1. In the above figure if ara 4B = 11 in,, arc 
ord Ali^l iu., find chord A'B' without meaauring 


' = ljlii., and 


Ex. 2. Draw two circles so that the radius of oc 
the other. 


. i. 


the diameter 



Ex. 3. To which o£ the elassea of figures mentiousd iu Art. 16 
oes a sector ijelong f a segment i 



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the circle 111 

Proposition VI. Theorem 

220. In the same circle, or in equal circles, the greater 
of iv:o {minor) arcs is subtended by the greater chord; and. 
Conversely, the greater of two chords subtends the greater 
(minor) arc. 





Given the equal circles and O', and are AB > aro 1>F. 

To prove chord AB > chord DF, 

Proof. Draw the radii OA, OB, O'D, O'F. 

Then, in the A AOB and DO'F, 

OA=0'D, and OB^O'F. (Why?) 

Z is greater than Z 0', Art, 217. 

(for arc, AB > arc I)F, and, in the same G, or in equal 0, of two un- 
equal arcs the greater are subtends lite greater angle at the center) . 

.■. chord AB > chord DF, Art. lOT. 

(if !«■(> A have two sides of one eqitat to tiro siiles of the olhci-, but the 
included angle of the first greater, etc.). 

CoNVKRSELY. Given the eqiial eh-eles and (y , and 
chord AB > chord DP, 

To prove arc AB > are DF. 
Proof. In the A AOB and DO'F, 

OA = 0'D, and OB^O'F. (Why!) 

AB> BF. (Why?) 

.•. Z is gi-eater tlian Z 0'. (wiiy ?) 

.-. arc AB > are BF, Art. 2IT. 

(in the name O, or in ei/Ma! ®, of tiro viirqual central i llie grimier 

"™"" '"^"' e, £. B. 



tnyle intercepts the greater 



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112 



liOOK a. I'LASE (iKOME'lMlY 



Proposition VII. TiiEOREJi 

221. ,1 iiiameter perpendiculnr to a chord hi-wcls the 
cJwrd and the arcs subtended by the chord. 




Given tbe circle 0, and the diameter PQ X ch. 
To prove that PQ bisects the chord AB and the ai 
and -1 <JB. 

Proof. Draw the radii 0-4, and OB. 
Then, in tlic rt. A OAR and OBR, 
OA = OB. 
OB = OB. 
:. AOAB = A OBB. 
:. AE=BR. and ZAOB = I BOB. 
:. are .1 P. = arc HP, 
(in ihr .wmr O, n;- in = ©, = central A intercept = 



i-d AR. 
isAPB 



(Why?) 
(Why?) 
(Why?) 
(Why!) 
Art. 215. 
■<■/.)■ 



^AOQ ::=: ^ BOQ {-'^A 15). .\ HTQ AQ ^ ATC BQ. (Wbj t) 
Q.E. D. 

222. Con. 1. -i diameter which l/isecis a chord {shorter 
than a diameter) i« perpendicular to the chord. 

223. Cor. 2. The perpendicular bisector of a chord 
passes through the center of the circle, and bisects the arcs 
subtended hy the chord. 

224. Cor. '6. A line from the center perpendicular to 
a chord bisects the chord. 

225. Cor. 4. A line passing through the midpoints of 
a chord and Us arc passes through the center, and is a diame • 
ter perpendicular to the chord. 



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THE CIRCLE 113 

Proposition VIII. Theorem 

226. Ill the same circle, or in equal circles, equal chords 
are eqiudisiaitl from the center; and, conversely, chords 
which are equidisiani from the center _ are equal. 




Givea the cii-ele in wliich the eliords AB ami CD 
are equal. 

To prove that AB and CT) tire eqnidistant from the 
center. 

Proof. Let OE he drawn ± AB, and Of ± CD. 

Draw the radii OA nnd Of. 

Then OE bisects AB, and OF Wsects CD, Art. 224. 

(a Jiiicfriym the milcr ± n chord biser.ls the chord). 
Hence, in the rt. A OAE and OCF, 

OA = 00. (Why?) 

AE = GF. Ak. s. 

.-. AOAE ^ A OCF. (wiiy?i 

.-, OE = OF. (Whj?) 

Conversely. Given circle 0, and AB and CD equidis- 
tant from tlie (tenter. 

To prove AB=CD. 

Proof. Let the pupil supply thu proof. 



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114 POOK II. l'LA>JE GEOMETRY 

Proposition IX, Theorem 

227. In the same circle, or in equal circles, if two chords 
are unequal, they are uHequally distant from the center, and 
the less chord is at the greater distance from the center. 




Given in the circle the chord CD < chord AB. 
To prove that chord CD is at a greater distance from the 
center than chord AB. 

Proof. Let OG be drawn 1 CD, and OF X AB. 
Then chord AB > chord CD. Hyp. 

.-. arc AB > arc CD, Art. 220, 

(iH the sameO, or in eijml. ®, Ihe greater of two minor arcs is svbtendiil 
h}i Ihe ijrcat'.T clwrd, mid coiwcrscly) . 

Mark off on the arc AB the arc AE^&n: CD, and draw 
the chord AE. 

Chord AE = chord CD, Art. 219. 

(t» the same O, of in equal ® , equal arcs are subt/ntiled by equal cltords). 
Let OS he drawn ± AE, and intersecting AB at L. 

.: OS = OG, Art. 226. 

(in the same O, or in equal ®, equal chords are equidistant from the 

But 0S> OL. Ai. 7. 

Also 0L> OF. (Why!] 

Much naore then OS, or its equal OG > OF. Ax. 12. 

Q. B D. 



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THE CIRCLE 115 

pRorosiTiON X. Theoeem (Converse op Peop. IX) 

228. Jn the same circle., or in equal circles, if iwo chords 
are nneqimlly distant from the center, the more remote is 
the less. 



Given in t!iR ciriilc llie olioi-d VI> farther from the 
center thao tlio chord All. 

To prove ehunl CD < cJioiil J /.'. 

Proof. Let OH be (iiwwn 1 CTi, mul OG ± AB. 

OH > CXr. (Why?) 

Oil OH mark off <>L = 00. 

Through L let the eliord ELF he drawn X OE. 

Then chord EF = chord AB, Art. 22G. 

(in Hie same ,C 

But the are CD < are EF. As. 7. 

.-. chord CD < ehord EF, or its equal AB, Art. 220. 

(in (fte same Q, or i« fqnn/ ®, ilie greater of tico minor arcs is sublended 
by ilic greater chord, and conversely-), 

Q. £. 9. 



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lit; BOOK IT. TLAXE GE03IETK1' 

PliOI'OSlTION XI. TUEOKEJI 

229. A KiraUjht line pcrfeiuUcular to a nidiits at Us 
fxtreinitii is Unnjint to tiic circle. 




Given the circle 0, tlie i-adiua OA, and \hc line HC ± OA 
at its extremity A. 

To prove that BC is tangent to the . 

Proof. Take P, any point on the line BC except A, 
and draw OP. 

Then 0/' > OA. (Why!) 

Hence the point P lies witliont llie circle. 
.'. every point in the line BC, except A, lies ontside 
the O. 

,■. BGig, tangent to the circlf, Ari. 2Ci. 

(« (ann^^il toaO is a straight hi,,- ichkk, etc.), 

0. E. B, 

230. Cor. 1. The radius drami to the point of contact 
is perpendicular io a tangent to a circle. 

231. Cor, 2. A perpendicular to a tangent at the point 
of cantad passes through the center of the circle. 

232. Cor, 3. The ■perpendicular drawn from the center 
of a circle to a tungent passes through (he point of contact. 



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THE CIRCLE 117 

pROPOSiTioK xn, Theorem 
233. Two parallel lines intercept equal arcs on a eir- 




Case I. Given AB (Fi<r. 1) fcangeat to tlie G I'CD at 
P, CD it sGiJiuit II AB iiiid iuterseGtiug the circumfciuiieo in C 
andi>. 

To prove arc PC = are PD. 
Proof. Draw the diameter PQ. 

Then PQ ± AB, AH. 230. 

PQ 1. CD. Art. 123. 

.-. are PC = are PD, Art. 221. 

(a diameter X chord iisR./s (/ib diw.f iKiiJ the ai-e« stMendea by 

the chord). 

Case II. Given .iB and CD {Fig. 2) [| seciiiits inter- 
secting the circumference in A, B and C, D respeefcivoly, 

Tb prove arc AQ — are BD. 

Proof. Lota tiLiigent ^i^ he drawn |UIB unLltouehing tlie 
circle at P. 

Then EF\\ CD. (Wliy?) 

Then arc AP = arc BP. Case I, 

:^ CP= arc DP, CVVhy!) 



Also 
Hence 



arc AC - 



iBIJ. 



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118 IKJOK II. rLAXK CEOMETRY 

Case III. Given Ali and CI) (Fig. 3) \\ t;uigeiits toucli. 
iag the O at P and Q respectively. 
To prove arc PEQ = are I'FQ. 
Proof. Let the tmpil supply the proof. 

234. Cor. The straight line whk-li joins the points of 
contact of two 2}ctraUel tangmiix is a dinmeier. 



Peoposition XTIT. Theorem 



285. Through three points, not in the sa 
%e circumference, and only one, can be ih-a 




Given A, B am! C any three points not in the same 
straight line. 

To prove that one eirenmference, and only one, can be 
drawn through A, B and C. 

Proof. Draw the straight lines .4^ and BC, and let _b 
be erected at the midpoints, D and E, of AB and BG 
respectively. 

These -li will intersect at some point 0, Aft. vi'i. 

{lines ± nmi-paralUl lines are not || ). 

But is in the X bisector of AB. Const. 



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THE CIRCI,E 119 

.*. is equidistant from the points A aud B. Act. H3. 
In like manner, is in tlio J. bisector of BC, and is 
equidistant from the points B and C. 

Hence ia equidistant from the three points A, B and G 

Henee if a circumference be described with as a cen- 
ter and OA as a radius, it will pass through .4., B and 0. 

Also DO and EO intersect in but one point. Art. e4. 

Hence there is but one center. 

Again, is equally distant from the points A, B and C; 
henee there is but ojte radius. 

With only one center and only one nuliua, but one (lir- 
cumference can be described. 

Hence one circumference, and only one, can be drawn 
through the points A, B and C. 

Q. E. B. 

236. Note. The theorem of Art. 235 enables ua to shrink or 
economize a circle into three poiiits ; or to expand any three points 
into a circle. 

Ex. 1. How many eircurafereneea can be passed through fouf 
given points in a pliine, each cireumferenee passing tbrou;i:h three, 
and only three, of the given points I 

Ex. 2. Draw two eirclea so that they outi have a common chord, 
Ex. 3. Can two circles whii-h are tangent to each other have a. 
common chord ! 

Ex. 4, Can two circles which are tangent to each other have s, 
common secant 1 

Ex. 5. Draw two circles which can have neither a 
nor a common tangent. 

Ex. 6. Is it poBsibie to draw two civulus which 
common secant f 



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I'LANE GKOMET: 



Proposition XIV. Theorem 



237. The two iangenis drawn io a circle from 
outside the. circle are cqiinl, a>id make, equal (o/y/cs 
line drawn from the point to the center. 



point 
icith a 




Given PA and PB two tangents drawn from the point 
P to the circle 0. 

To prove PA^PB, and ZAPO= ZBPO. 
Proof. Let the pupil supply the proof. 

238. Dbf. The line of centers of two circles is the 
line joining their eentei's, 

239. DBF. Two circles which do not m(!ct m»y have 
four common tangents. 

A commoa internal taageat of two circle.'^ is a tangent 
which cuts their line of centers. 

240. Dep. a common external tangent of iwo circles is 
3. tangent which does not cut their line of centers. 

Kx. 1, In the above Sgure, prove that the line ilrawu to the center 
from the point in which the two tangents meet makes equal aiiglsB 
with the radii to the points of eontaet. 

Ex. 2. Draw a circle with a radius of 3 ia. and anothei' with a radius 
of 3 in., with their oeutera 4 in. apart. "Will these eirelea intersect T 
n their centers were 6 in. apart, would they intersect S 



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THF, CIRCLE l^J 

Proposition XV. Theorem 

241. Iftico circles intersect, their line of centers is per- 
pendicitlaf to their common chord at its middle point. 




Given the circles ami 0', itilevscetinK at tlm points 
L and B. 

To prove 00' ± AB at its middle point. 
Proof, Draw the radii OA, OB, O'A, O'B. 
Theu OA^OB, und 0'A = 0'B. (Why?) 

Hence and 0' are two points each equidistant fi'oin 
1 and B. 

:. 00' is tlie ± bisector of AB. ah. in. 



Braw two cirules in which tlie Una of e 
Ex. 2. Equals the sum of the lailii. 



L and />. 



.-. 00' is tlic X bisector of AB. Avt. 113. 

Q. E. D. 



Draw two eireles in which tha 1 



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V22 liOOK ir. I'LANK (lEOMETKY 

PiiOPOhiiTioN XVI. Theorem 

242. Jf two circles are iaiujeut to each oiker, fke Uite of 
eenlers passes (hrough the point of contact. 




Given tho cirdes iiml 0' tangfiiit to each otliev at tbe 
poiut R. 

To prove that the line 00' passes through It. 
Proof. At the point E let PQ, a tangent to the given 
®, be drawn. 

Also let AB be drawn 1 PQ at R. 
Then AB passes through and also throngh 0', Art. 231, 
(a l-loa tangeni al the fioint of coiilact pusses Ihrough the center). 
:. line AB eoincides with the line 00'. Art. 64. 

.'. 00' passes through the point E, 
(Jor it coiittidm ititli AB ahieh passes through li ) . 



How many eomnioil interail, and 
tangents bave two eirolea 
Ex. 1. If they toneh esternally t 
Ex. 2. If they touch iDternally f 
E*. 3. It they intersect ! 
Ex. 4. If one circle lies wholly with 
Ei. 5, If one circle lies wholly witl 



external 



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THE CIRCLE 



eXERCISES. CROUP I 



Ex. 1. The line joining the oenlor 
cKord is perpeadienlar to the chord. 



I four points t alt en i 



'O is tho perpendieul 
1' upon two clioi'ds a 



[■0 BD; prove that chord 



Ex. 2. A, B, Cand D n. 
eircumferenee of a eirele, 
AC=ahovii BD. 

Ex. 3. Taugeuts drawn at the estron 
p&ratlel. 

Ei. 4. PA and FB aro tangents to a eii 
P. is the center of the oirele. Prove th; 
bisector o£ the chord AB. 

Ei. 5, If the perpendiculars from the Ci 
equal, tlio aroB subtended by these chords a 

Kl. 6. A, B. C and D are points taken i 
onmfecence, and arc AC is greatei' than t 
AB> chord CD. 

Ex. 7. State the converse of the preceding 
theorem and prove it. 

Ex. 8. Given EA, EQ and QB tangents o£ 
the circle ; prove Rq=EA + QB. 

Ex. 9. If a quadrilateral be e ire urns cribed 
about a. eirele, show that the eum of one pair of 
opposite aides equals the auiu of the other pair. 

Ex. 10. if a hexagon be eireumscribedabout aairolo, show that tho 
aiim of three alternate aides equals the sum oC the other throe sides. 

Ex.11. If a polygon of 2rt sides 
(he sum of n alternate sides equals t: 

Ex. 12 Acm 
inequilateral, 

Ex. 13, If two circles b 
externally, the eoiuiuuii intei 
bisects the common external ti 
prove PA = FB). 







ribed paralleiograi 



mgent 



.tl,i., 




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il;4 



BOOK II. I'L.VNR nKOMF.TliV 



Ex. 14. It two circles are tniigfi 
(either e.tteniiilly or iiitei-nally) taugcti 
drawn to llieiu from any point iu tie eon 
mon tangent ara equnl. 

Ex. 15. Two eircli 



O' are tangent mternnJly 
at P. The line PAB ia 
drawn intersecting the 
Prove that OA and O'B 





MEASUREMENT. RATIO. 

243. Measurement. For many purposes, the most; 
advantageous way of ilealiog with a given magnitude is to 
take a certain definite part of the magnitude as a unit, 
and to determine the nnmbei' of times this unit must be 
taken in order to make up the given magnitude. Ease and 
precision iu dealing with magnitudes are thus obtained. 

Geometric magnitudes thus far have been treated aa wholes, the 
object being simply to determine whether two given magnitudes are 
equal, or unequal, or to determine some similar general relation. 
Hereafter geometric magnitudes will frequently be treated as if com- 
posed of units. 

To measure a given magnitude is to iind how many times 
the given magnitude contains another magnitude of tlie same 
kind taken as a unit. 

244. Tlie numerical measure of a magtiituJe is the num- 
ber which expresses how many times the unit of measure is 
contained in the given magnitude. 

Thns, when a boy says that he is five feet tall, he means that, if afoot 
rule be applieti to Lis height, the foot ruie will be contained live times. 

A quantity is often measured to beat advautage by measuring a 
related but more accessible quantity, which Las the same numerical meas- 
ure as the original quantity. This process is indirect measurement. Thus, 
the temperature of the air is measured iniiirettly by measuring the 
height of a column of mercury in a thermometer tube. So the number of 
times a unit of angle ia contained in a given angle is often ascertained 



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MEASUKKMKKT 



125 



moat readily by determining tbe number of times a unit of arc is cod- 

taiued ttt a give □ a. re. 

245. The ratio of two niiigniUKius of the same kind is 
their relative magiiitatle as determined by the number of 
times one is contaioed in tbe other. Hence, it is the qiio- 
tient, or indicated quotient, of tlie two miignitudos. 

Thus, tbe ratio of 3 ft. to I ft. 7 in. ia ^~. <"■ fg- 

Tlie use of ratio U illustrated by tUe fuct tbat several indicated quo- 

tiimts wbcil taken togetlier may be simplified by canoellation before a 

floal determination of tbeir vnlins La iiinde. 

Two magnitudes of the same kind have tlie same ratio as 

their numerical measures. 

246. Measurement as a ratio. Au important piirticular 
instance of ratio is tbat ratio in wliicb one of tbe two mag- 
nitudes compared is a unit of measurement. Hence, the 
nnmerical measure of a magnitude is the ratio of tbe mag- 
nitude to tlie unit of measure. 

Tims, tlie numerical meaaure of tbe beiglit of a boy is tbe ratio of, 
bisIieigU(5 ft.) to tbe uait of measiiru (I ft.), orS. 

247. Commensurable maguitudes are magnitudes of the 
same kiud which have a common unit of measure, 

Tiius, 12 ft. and 25 ft. bave tbe common unit of measure, 1 ft., and 
ience are commensurable magnitudes ; also 13! bu3, and 7J biia, liava 
a common unit, I peclc, and are commensurable, 

248. Incommensurable magnitudes are magnitudes of 
the same kind wliich have no common unit of measure. 

Ills, $5 and Iv'a ; 7 vrs. and -^15 yrs.; so tbe side and the diagonal 
of a squHve may be proved to bave no common unit of measure ; likewise 
tbe diameter and tlie circumference of a circle. 

In general, a ratio wliicb is expressed bj a surd number, 
^\/S or\ 3, is a ratio between incommensurable magnitudes 
(called au incommensurable ratio). 



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'I,ANE GEOMETRY 



METHOD OF LIMITS 



a variable, is fi qnanlity 
imlK'f of differfijit valnes 

or |3voi)ltMii . 
goes varies with the nurabei of 



tou w 


h 










250 


\ constant 




u 


ivhicli remains iincliiiiigeLl 


in \ 


u a 








Lissioii. 


 h 




g 






circle and tlie iiuinUcr of sides 


oftb 










itc, the pariiiietet of tUe pnly- 


j;OI\ 




e 1) 






iioe of tlie circle nill remain 


uncLan^ 


an 


li b 




u 




251 


T 


mto 




JJ 


mtity is a conatiiut quantity 
^fiven discussion, the given 


\-M 











5 we please in Talue, but 



1 the circumference of tlio circle 

]a til lie n e bed i>olygun ; also,- the area of 

fUe circle is the limit of the urea of the inscribed polygon. 

From the definition of limit, it follows that the differ- 
finee between a variablo and its limit may be made as small 
as we please but can not become zero. 

As another illustration of a variable and its limit, we may talie the 
ease of a point Ptravelling along a given line AB, in auch ft way that 
in the first second it passes over APi, one-half the line AB; in the 
second second over half the remaining part o£ the line, and arrives it 
P;] in tlia third second over one- 
half the remainder of the line, and , , , , 

arrives at Pj, etc. It is evident -^ fi ^' ^^ ^ 

that the point P can never arrive at 

B; for, in order to do this, in some one second the point would need 
to pass over the whole of the remaining dietanee. 

In this illustration, AP, the distance traveled by the moving point, is 
a variable (depending on the number of aeconds), and AB is its limit. 

If the distance AB be denoted by 2, the distance traveled will be 
denoted by 1 -fl + i -|- i + , . . . and will vary according to the 
number of terma o! the series taken. 



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SfETHOD OF LIMIT3 



127 



252. Use of variables aad limits. Many of the proper- 
ties of limits are the same as tlie properties of the variables 
approaching them. Ileuce a demonstration of a difticiilfc 
theorem may often bo obtained by first fintling the properties 
of relatively simple variables and then transferring these 
properties to their more complex limits. 

253. Properties of variables and limits. 

1. The liinit of iJte sum of a nitmbfr of eari^lfS equals 
the sum of the limits of these vaHables. For, since the 
difference between each variable and its liinit may be made 
as small as we please, the sum of alt these differences may 
be made as small as we please (since it is a finite number 
of differences, with each difference approaching zero). 

2. The litnit of a times a variable equals a iiiiies the 
limit of the variable, a being a constant. For, if the differ- 
ence between a variable and its limit may be made as sinall 
as we please, a times this difference may be made as small 
as we please. 

3. The limit of -tk part of a variable *■'< -fh part of the 
limit of the variable, a being a constant. For, if the 
difference between a viiriable and its limit may be made an 
small as we please, — th part of tliis difference may be made 
aa email as we please. 

4. If a variable ~0, a times (a being finite) or - part of 
the variable ~0; and if A diminish in the one, or i 
the other process, the limit is sliU zero. 



Ez, 1. Are 2) gal, mid 3^ qt. c 
Ei. 2. If c donotc u miistaiit bi 

eacU ot the following a v&riable or a 

Uftte. 



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LASE (iEOJlETRY 



XVII. Tin:f)iu.:M 



254. 7/ tu:> rariabU^ nr- .iJmnj^ pqml, and mch < 
.■or.clu^.i a UntU, /luir Umil^ un- 'qiuil. 




y I> <- 



GiTen AB flic limit of the variable Al', AC the Ihuit of 
the variable AQ, and AP^AQ always. 

To prove AB = At7. 

Proof. If the limit AB does not equal the limit AC, 
one of these limits, as AG, must be larger tiiau the other. 

Then, on AC, take AD equal to AJi. 

But AQ may have a value greater than AD. Art. U.jl, 

Hence AQ would be greater than .i/>, 
Or AQ > AD tt-hi.)h = AB) . 

. AQ > AP, 



{Jor 



AP). 



But this is eontrary to the hypothesis that AQ aud AP 
are always equal. 

,'. AC cannot be greater than AB. 
Id like manner it may be shown that AB is not greater 
than AC 

:. AB^AC. 



Ex, "Wliiit method of proof is used in Prop, XVII! 



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MEASUREMENT OP ANGLES 



Proposition XVIII. Theorem 



255. In the same circle, or in equal circles, ix-o 
angles have the same ratio as their intercepted arcs. 



Case I. T17jfw the intercepted a 



tral 
'mmensuraile. 





To prove 



Given thuoqiuil® C and 0, with tlie ceutrai ^A'O'B' ami 
AOB intercepting the commensurable arcs A'B' and AB. 
ZA'O'B' ^A'B' 
lAOB AB' 
Proof. Let arc PQ {from n circle - ami 0') be ix com- 
mon moasiiro o£ tho arcs A'B' and AB. 

PQ will be contained in arc A'B' an exant number of 
times, as 7 times, and in AB an exact number of times, 
as 5 times. 

arc AB 5 

From (y and draw radii to the several points of 
division of the arcs A'B' and AB. 

Then the ZA'O'B' will be divided into 7, and the 
/.AOB into 5 small angles, all equal. Art. 216. 



{in the «!. 



Fj =® equal arcs mbtend eqviU A. at the t 
lA'O'B' ^1 
lAOB 5 
ZA'O'B' ^A'Ii' 
I AOB AB' 



Art. 245. 



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BOOR II. I'LAXE 



Cask II. inint the h<lr>rrpf''d an 
siirahle. 





Giventheeriiiii! ciiO'iimlO,witli the eeiiti'id A A'O'B' a.hA 
AOB iuteruepting the iuooramensorable arus A'B' and AJi. 



To prove 



ZA'O'B' _ 
' lAOli 



AlV 
AH ' 



Proof. Let tlie arc AB be divided into any naniber of 
equal parts, and let ooe of these parts be applied to the 
arc A'B'. It will be contained in A'B' a certain number 
of times, with an arc BB' as a remainder. 

Henoe the arcs A'D and AB have a common unit of mess- 



lA'O'D A'D 
  lAOl) AB ' 


Case I. 


If now wu let tlie unit of measure be iiidcflii 


ilelydimin- 


ished, the arc DB', which is less thun tlie unit of i 


ncasuro.will 


be indefinitely diminished. 




Hence arc ^'i>=^arc A'B' a-s a limit, and 


I A'O'D- 


^1^'0'5'asalimit. 


An 351. 


Hence  , . ..„ becomes a vafiable approaohi 


ZA'O'B' 
" lAOB 


as its limit; 


Alt. 253,3. 


Also -j^ becomes a variable approaching 


A'B' . 


limit. 


Art. 253, 8. 



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MKASri!EMENT OF ANGLER 131 

Lsel. 



256. Dep. a degree of arc is one tliree luiiiJf.'d and 
sixtieth part of the eircura fere nee of h (circle. 

257, Cor. TJie, mcniber of degrees in a cfnttul umjle 
equals the niimher of degrees in /he hitercepled an: ; thut is, 
a central angle is measiirrd by Ua Intercep/ed are. 



Ex. 1. What is the ratio ot a quadrant to a semi-o Ira u inference ! 

Ex. 2. "What is the ratio of an angle of an equilateral triangle to 
ne of the acute angles of an isoaeeles riglit triangle t 

ot each circle is on the 



Ex. 4. Draw three circles so that the center o£ each ia on 
camterenee of the other two. 

[SuQ. First draw an equilateral triangle.] 

Ek. B. Draw three circles each of which shall he tangei 
other two. 

Ex. 6, Draw tv 
to one ot these cirt 

Ex. 7. Draw two concent rit 
one and a chord o£ the other. 



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liOOK II. PLANE GEOME 



Proposition XIX. Tbeork.u 

258. An inscribed angle, is measured by one half ifg 
Inlercepted arc- 




ana AB 



(Whyf 
(Whyfl 



Case I, When the n-nl 
the inscribed angle. 

Given Z2J.4C(Fig. 1) iusei-ibecl iii the I'ircle C 
passing through the center 0. 

To prove that Z BAC is measured by * aw. BC 
Proof. Draw the radius OC. 
Then, in the A OAC, OA = OC. 
I A = IG. 
But lIiOC= ^A +Z0. 

.-. ZB0C = 2Z.-1. 
But Z HOC is measured by arc RC, 

{aceitlral Z iiineo-mredhmlsixlcrccincd arc). 
.'. Ko. of iingulur degrees in / .BOC^No. of arc dfgi 

Hence Z -i is iiieusnred by \ arc SC. Ai. 5. 

Case II. Wheii the center of the circle lies within the 
ht scribed tmgle. 

Given the inscribed /IBAG (Fig. 2), with the eentec 
of the circle tying within the angle. 

To prove that /.BAC is measured by h arc BO. 
Proof. Draw the diameter AD, 

Then /.BAD is measured by ^ are BD. Caael. 

Let the pupii complete the proof. 



aBC. 



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MEASUREMENT OF ANGLES 



133 



Case III. When (he cenier of the circle is witliout the 
inscribed angle. 

Given the inscribed IBAG (Pig. 3) with tlie center 
outside the angle. 

To prove that ABAC is measured by h arc BC. 

Proof. Let the pupil supply the proof, 

259. Note. By use of the nbove theorem, if the number of 
degree* in the intereepfed arc be known, the number of degrees in 
the insoribed Bnslo enn be determined immediately. Thus, if the arc 
BC (Pig. 3) eoQtaixiS 4S°, the angle BAC contama 24°; also, if it be 
known that the mslo BJC e< 




260. CoE. 1. AU angles inscribed in the same segment, 
or in equal scgmeni.t, are equal. 

Thns A A, A', A" (Fig. 4) are all equal, for each ot 
them is measured by one -half the arc BBC. 

261. Cob. 2. An angle inscribed in a semicircle is a 
right angle, for it is measured by one-half a semicircum- 
ferenee. 

Thus FE (Fig. 5) is a diameter .-. ZFGEis a rt. Z. 

262. Cor. 3. ^1k angle inscribed in a segment greater 
fhan a semicircle is an acute angle; an angle inscribed in a 
Segment Jess than a semicirch is an ohttise angle. 



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iS-l BOOK II. PLANE CEOMI'mji: 

Proposition XX. Theoreji 

263. An angle formed bi/ two diordn Intersecting leiihin 
a cireiimfiirence is medsurrd hij half the sum of the inter- 
cepted arcs. 




Given tlic chords AB and CD in the G AVBC, inter- 
secting within the circiimferenco at''tlie point P. 

To prove that /.DFB is measured by J (arc BB-\- ai-o 
AC). 

Proof. Draw the i^hord AB. 
Then, in A ABB, 

IDBB= IA+ IT), Art. 13S. 

[an exi. I of a £\ is equal lo the sum of the lu'o opp. int. i ). 

But Z J. is measured by i are DB, Art. 358. 

{an inscribed Z is measured l/ij one-half its mlcrvepte/i are). 

Also Z 7> is measured by i are AC. (Why!) 

Hence I DFB is measured by A (arc JJB + arc AC). 

Ax. 3. 
Q- E. B. 

Ex. 1. lu the above figure, i! arc />/( eonUins W and are -IC coa- 
t*ms 38°, how many degrees in lAPC * ia lAPD ? 

Eic. 2. If arc AD = 5i;° and Z AFD = 124°, fiud are CB. 

Ex, 3. If, in Fig. I, p. 132, arc AC oofttalna 112°, how many 
degrees are there in tiie I A ! 



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MEASUREMENT OF ANGLES 135 

Pedposition XXI. Theorem 

264. An angle formed by a tangent and a chord drawn 
from the point of contact is mfamred by half the inter' 
cepied arc. 



Given the O PCD, awl Z.APC formed by the tangent 
APB and the chord PC. 

To prove that Z APC is measured by i are PEC. 

Proof. Let the ehord CD be drawn ll AB. 



Thtii 


Z.APC -^ 


■- I PCD. 




(Why?) 


But 


I PCI) is maasu 


red by I a 


.i-c PD. 


(Why!) 




.-. I APC is measu 


red by i a 


re PD. 


Ax. 8. 


But 
(i. 


arc PEC = 
00 W Uiie,^ inkrcept cijuw 


 arc DP, 


■rcumfermce). 


Art. 233. 



,■, /APC is measured by i arc PEC. Ax, 8, 

Also Z.BPC, the supplenieut of ^APC, is measured 
by i are PDC, Ax. 3. 

(for urc I-DC is the conjmatc of arc I'liC) . 

Q. E, B. 

Ex. 1. In the iibove figure, if are /'A'(,* uoKtaius l::4'^, Low many 
degrBea lire in the angle AI'C ! 

Ei. 2. If aru CD = 'M", find the angles on the figure. 



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rKorosmoN XXJT. Thkorrjc 

265. An an'jle formed by two secants, or hij lirn tangents, 
or hi! a secant ami a tangent vieel in;j withmil the circumference 
is measured by half the ilifferntre nf Hip hilerrrjiled ores. 



FiS.t 3?is.%. (fib', a 

I. Given the O ACDB (Fi^. 1), and the lAPB formed 
hy the two secants PA and P£, meeting at the point P with- 
out the circumference. 

To prove that ZP is raeasnrerl hy 1 (arc>4B— arc CD). 
Proof. Draw tlie chord CB. 

Then IACB= ZP+ZB- [Whj-i) 

.■. ZP= ZACB— ZB. Ax.3, 

But ZACB is measured by i arc AB. Art. 258. 

Also IB is measured by 4 arc CD. (Why!) 

.•. /P is measured by A (arc ^P — arc CD) . Ax.s. 

II. Given the O ABB (Fig. 2), and ZCPB formed 
by the tangent PC and the secant PP. 

To prove that Z P is measured by * (arc ABB — Are AD). 
Proof. Let the pupil supply the proof. 



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KXKliCTSF.S OK THE CTRCLK 137 

m. Given Z^Pfi (Fig. 3) formed by tlie tangents PC 
and PD. 

To prove that ZP is measured hy i {arc AEB — 
are AFB) . 

Proof. Let tho i>iipil supply the proof. 

266. Note. By means of Props. XVIII-XXII, angles formed 
by chords, seeanta, or tangents, or oombinations of these, are all 
reduced to eentrp.1 angles and hence may readily be compared. 

Ex. 1. If, in Tig. 1, p, 130, are CI> = Zi° and arc .JJV = lOS", 
draw AD and find all the angles of the figure. 

Ex. 2. If, in Fii-. 2, p. 13U, angle i' = SO'', find arcs-Ji'fl and AEB. 

EXERCISES. CROUP tS 

Ex. 1. What now methods of proriug two lines equal are afforded 
by Book ir t 

Ex. 2. Wliat new methods of proving two angles equal '. of prov- 
ing two angk-5 supplementary f of proving an angle a right angle 1 

Ex, 3. What methods of proving two arcs equal ! 

Ex. 4. In ii quiidrilareral ioscribed in a circle, tiicli pair of uppiwiti; 
angles is supplcinciitiiry. 

Ex. 5. A «hord forms equal angle? with the tangents at ita 
extremities , 

Ex. 6. If an isosceles triangle be inscribed in a circle, the tangent 
at its vertex makes equal angles with two of its sides and is parallel 
to the third side. (Is the converse of this theorem inic ?) 

Ex. 7. I£ two chords in a circle intersect within the i^irele at right 
angles, the sum of a pair of alternate area equals 
a semioircumf erenee . 

Ex. 8. Given the center of a circle and 
-iC a tangent ; prove IBAC=X<<^0. 

Ex. 9. If one side of an inseribed quadri- 
lateral be produced, the exterior angle fio 
formed eiiuals llio opposile interior hi 




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i;;s 



];ooK II. ]']^ANE ^.EO^rKTuy 



Ex. 10. IE twi> tiingoiits (o a. oirele incluiio an nn^iia of 60" at 
their point of intersection, the chord joining the ]ioiut3 of eontaot 
forms with the tnngenta an equilateral tl'iangle. 

Ex. 11. Tliu chord A/I equals the eijord C]) lu a given eirele, and 
the chords, if prodneoii, iutei'svet at the point i'. Prove seeant PA^s 
secant I'C 



pribed ahout the triani^le ABC, and P ia 
Prove that the angle Jill' equals oue- 



Es. 12. A circle iE 
the uiidpoint of the a 
half the angle C. 

Kk. 13. It A, n, C, 1> and E be poiTits taken ii 
eireumtercnee of a eirele, and the area AU, BC, CIJ and /IE be equal, 
prove that the angles AliC, HCI) and CDl! are equal. 

Ex. 14. An inscribed angle formed by a diameter and a chord baa 
its intercepted are biaeeted by a radius which ia ,, 

parallel to the chord. 

Ex. 15. Two see^mts, PAI! and PCD, intersect 
the eirele ABDC. Prove that the trian>;les P/IC 
and rj/>are mutnally equiangular. 

Ei. 16. Given AC s, tangent ami An\\ri:,- 
prove A ACD and AJIE rautnally equiangular. ' 

Ek. 17. Given ABC an inaoribed triangle, AE 
± BC, and CD X AB ; prove arc iJO=arc BE. 

Ex 18. If the diagonals of an iuseribed quad- 
rilateral are diameters, the quadrilateral is a 

Ex. 19. Tangents through the voitie. 
inacribeii I'eptangle form a rhombus. 

Ex. 20. Two circles intersect at /' 
and Q. PA and Pli are diameters. 
Prove that <^A and QD form a straight 
line. 

Ex. 21. Two eqi 
at Pand Q, through P a line ia dra 
terminated by the eireumfersooes 
equals QB. 

[SuQ. i A and B are measured by what arcs !] 




J and B. Show that QA 



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EXERCISES. AUXILIARY LIXEK 



139 



267. Use of auxiliary lines. In deraoustfatiii^ theorems 
relating to tlie cifcie, it is often helpful to draw one or 
more of the following; auxiliary Hues: A radius , a diame- 
ter, a chord, a perpeirdicuhtr from the coikr upon a chord, 
an arc, a circtimferfiu-i^, etc. 

EXERCISES. CROUP 17 

AC'XTI.IARY LIXKS 



•J ; prove BM = r.X. 

Ex. 2. If from any point in the eircuraferenee 
o! a circle a eliord and a tangent be drawn, llic 
perpendiculars drawn to tLem from the midpoint 
of the are subtended by the chord are eq^ual. 

Ex. 3. Given the center of a eirule, and 0J> 
X chord AC ; prove IA0D= IB. 

Ex. 4. Prom tbc extremity of a cHametcr chords 
are drawn making «iual angles ivitli tiie diameter. 
Prove that these chords are equal. 

ISvG. Draw Ji from the center to the chords, et 
la the converse of this theorem true ? 




Ex. 5. If through any point within a circle equal chords he drawn, 
show that the line drawn from (he center to the point of intersection 
of the chords hisects their angle of intersection . 

Ex. 6, Tangents /'J and W are drawn to a cir 
O. Prove that angle i' equals twiue angle OAIl. 



Ex. 7. If in B 

the segments of o 
other chord, 

Ex. 8. A paru 
rectangle, 

CSuo. 
P, 100.] 



le Iwp equal chorda ir 
orii equal the segment 



( the diajjouiils o£ thu i—/ , 




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rr,ANF, riT'O.MKTRY 



Ex. 9. If a qundvLlatera! Ije o ire am scribed nboiit a circle, Um anglea 
subtended at the center by a pair of opposite aidfis are supplpinentacy, 
[Sua, Draw radii to tlie poiiifce of contact and show that tiiere are 
four pairs of equal A at Hie center.] 

Ex. 10. If an equilateral triangle ABC be inscribed in a oirele and 
any point P be taken in the arc AB, show that PC= PA+FB. 

[SUG. On PC take Pif equal to FA, draw AM and prove ii PAR 
and MJC equal.] 

Ex. II. Two radii perpendicular to each otlier are produced to 
intersect a tangent, and from the points of intersection othei' tangents 
are drawn to the circle. Prove that the tangents last drawn arg 
parallel . 

[Sun. Draw radii to the three points of con- A 

tact, etc.] 

Ex. 12. A straight line intersects two oonot 
trie circles. Show that the segments of the I. 
intereepted between the eirpumferences are eqi 
(prove --15 = 01)). 

Ex. 13. A common tangent in 
drawn to two circles wliich ara 
exterior to each other. Show that 
the chords drawn from the points 
of fangenoy to the points where 
the line of centers cuts the cir- 
cumferences are parallel. 

Ex. 14. A circle is described on the radius of a 
diameter, and a chord of the larger circle is drawn from tlie point of 
coataet of the two circles. Prove that this chord i 
circumference of the smaller circle. 

[SuG. If the chord is bisected, a X from the c( 
the chord will also bisect ti:B chord, i 

Ex. 15. Two circles are tangent ex- 
ternally at the point P. Tiirough F any 
two lines APB and CPD are drawn ter- 
minated by the circumferences. Show 
that the chords AC and BD are parallel. 
[Sl'O. Draw the common tangent at 
F, If AC and BD are |1 , what A mui 




■ted hythe 
of the larger to- 




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EXERCISES. MAXIMA AND MINIMA 

Ex. 16. Two circles intersert at the points P Biid Q. 
and CQU uro drawn, terminated by the oiroumferences. 
^Cand BDare parallel. 

Ex. 17. If a square be described on the hypotenui 
triangle, a line drawn from the center of the square to 
the right angle bisects the rigrht angle, 

[SuG. Describe a circumference o 
triangle as a diameter.] 

Ex. 18. If two cirelea are tangent 
tangeut touches them at A and B, re 
right angle. 

Ex. 19. If in the triangle ABC th 
drawn, the augle ASD equals the angle AEB. 

[Sua. Describe a semieireumference on A 



righL 



1 the hypoter 



ixternally at !', and : 
pectively, the angle 



altitudes JIV and AE a 



diameter.] 

268. Def. a maximum is tlie greatest of a class of 
mawiiitudes satisfying eertalE given conditions, and a 
minimum is the least (see Arts. 470, 471) . For instance, of 
the chords in a given circle, which is the maximum 1 



EXERCISES. CROUP 18 



MAXDIA AND 



Ex. 1. 0! the chorda drawn through 
determine which is the greatest, and all 

Ex. 2. Find the shortest line, and 
also the longestline, that can be drawn 
from a given external point to the eir- 
eumterenee of a circle. 

[Sl'G. O beinp the center, prove 
in A PCO, PA < PC ; by A PDO, 
PB > PD.] 

Ex. 3. Find the shortest line, and al 
est line, that can be drawn to the cii 
of a circle from a point within the circle. 

[SuQ. O being the center, prove, by 
A Ol'C, FB < PC. etc.] 




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1-i'J liOOK II. PLANE l.!E 

Ex. 4. If two cirples intersept, show 

tU»l, of Hties drawn throufrli a piiict of , 

iJii-'ifei^tifiu and fenainnted by ihe> cir- 

(tumfereiitt'S, that line is a loaximiim 

which is pai'allpi to tlm 

liueof <^e»t6rs. 

[Sun. Prove US < 00' .: CPD < AI'B .] 
Ex. 5. Given AB 1 Oli in eirele O ; prove 
ZO.rBtbe ffiasimuraof all i having their vertieea 
im the oivcumferecee and their sidespasHing througli 
O and i! roepeetively. 
a eirelo on OA as a diameter.] 



EXERCISES. GROUP rs 

DE MO XST RATIO XS BY IXDIRECT MKTTIODS 
Prove the following by an indirect method (see Art. 195) ; 
Ex. 1. A segment of a circle which contains a right angle ia a 
semicircle. 

Ex. 2. If a rectangle be inscribed in a eirelo, its diagonals are 

Ex. 3. Prove the second part of Prop. VI by an indirect method, 

Ex. 4. Prove Prop. X by an indirect method, 

Ex. 5. A straight line connecting the midpoint of a chord and the 
midpoint of the arc subtended by \\\e chord is perpendicular to the 
chord (use the method of eoineidenee). 

Ex. 6. A line joining the midpoints of two parallel chorda passes 
through the center. 

[Sl'g. Draw a J- to each chord from the center and show that 

Ex. 7. If the opposite angles of a quadrilateral are supplementary, 
a circle oan be circumscribed about the quadrilateral, 

[SUQ. Pass a circumference through three vertices of the quadri- 
lateral; it it does not pass through the remaining vertex, etc.] 

Ex. 8. Prove the converse of Prop. XII, 



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EXERCISES . CONSTAKTS AND LOCI 



un 



269. A geometrical constant is n, gfom(;trteal inagiii- 
tude wliifih varies in some respect, as in position, but 
remains constant in size. Thus the angles inscribed in a 
given semicircle vary in position but are alt of the same 
size; viz., a right angle (see Art. 261). 



EXERCISES. CROUP 30 



XATIOX OF CON'STAXTf^ 



AXD LOCI 
Pis any point 



Ex. i. All and AC are tangents to a eirck 
cirouralerenee outaids tlie triangle ABC. As 
sura of tiie Z.-iand / JiPC is conetant. 

Ex. 2. In Es, 8, p. 12n, if AR and BQ be produced to meet at T. 
Bhow that tlie venmeter ot tho triangle Tffy equals tlie sum of TJ 
and TB, ami heiiee that the perimeter of triangle TBQ is constant, no 
matter how /' may vary in position betmeiin -J and IS. 



Es. 3, Shoiv o 



the 



■e that, i£ O 



i the c 



, Z.ROQU 






Ez. 4. Two circles intersect in the points 
A aod B. Prom any point P on one ciraum  
fereuee lines FAC and PBD are drawn, ter- 
minated by the other eireumference. Show 
that the chord CD is constant. 

[St-G. Draw BC and prove ICBD a, 



Ex. 6. Given OA J- OB, and Cn : 
(liven lungth moving so that D is alwa; 
and C in OH and 7' the niidjiohit c 
that OP is conf^faiit in IciikMi {-.c-e V 

The dftermmatioH of ha 
fucilitaied hij shoiviiig that 
magnitude is a conslunt. 

Es. 6. Find the locus ot a poi 
dlataiiCL* a troui a giveQ circumtei'i 




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144 isOOK II . I'LANE r.EOMETUY 

Ex. 7, Fiml Tho locus of Iho mUipoiiils ot the ni 



Ex. 8. Find the locus of tho mi.lnoiiits of :in chordsi of n ftiven 
leugtli di'awiL in a given eirole. 

Ex. 9. Find tlie )oi>us of tiie vertices of nU rij^lit lriaJii;li'K liaviag 
a giTen liypotouuse. 

Ex. 10. Find the loeus of the raidpointa of all tht ohoi'ds drawn 

[Sl'G. Draw a line from tho center to the given point and perpen- 
diculars from tbe teirtcr iipou fhn 
chords.] 



Ex. 12. QP 
length and 
always in a 
line. Find the 




EXERCISES. CnoyP 31 

THEOREMS PRO^TA) BY VAKIOTJS ^IKTHODS 

Ex. 1. The lino which liiseeta the angle formed by a tang 
a ehord biseefs the intercepted are also. 

Ex, 2. An inscribed trapezoid is isosceles. 

Ex. 3. Given TA and TI! tangents, are J7.'= T^ 
m", an Sll = s:,°, and are 1)0=150°; find all the 
angles of the figure. 

Ex. 4. If two tangents to a circle are parallel, 
the line joining their points of contact ia a diameter, 

[SuG. Draw radii to the points of contact and 
use an indirect method of proof,] 

Ex, 5. A rectangle cireiimaerihed about a circle ii 
[Sua. Use the preceding theorem.] 




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MIKCF.LLANEOUS EXERCISES 



Bx. 7. Find the angle formed by thi 
Bide o( an macrlbed square and the taugeii 
through the vertex of the square. 

Ex. 8. Prom an external point a seeant is drawn through the 
Center of a circle, and also two other secants making equal angles 
frith tbe first secant. Show that the seeanta last drawn are equal, 

Ex. 9. AliC ia a triangle and on the side AB the point P ii 
taken and on liC the point Q, so that angle Bl'Q equals angle C. 
Show that a circle may ba circumscvibed about the quadrilateral 

Ex. 10. Two oiiClBB are tangent to each other externally, and a 
line is drawn thtough the point of contact terminated by the circum- 
ferences. Show that the radii from the extremities of this line are 
PMallel. 

Ex. 11. If a 
Wangle as di^u 
triangle. 

Ex. 12. The chord of an are is parallel to the tangent at the 
midpoint of the are. 

Ex. 13. I£ a triangle be inscribed in a cirele, the sum n£ the 
angles inscribed ia the segments exterior to the trianijle is four right 
auBles. 

Ex, 14. Find the corresponding theorem for an inscribed 
"juadrilateral. 

Ex. 15. Find the loi^us of the centers of all circles passing through 
two given points. 

Ex. 16. If two unequal chords intersect in a circle, the greater 
chord makes the less angle with the diameter through the point of 
intersection of the chorils. 



e of this theorem. Is the 



Ex. 17. The .51 
tbe hypotenuse am 



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UG 



BOOK TI. PLANE GEOJtETKV 



touch the m- 
iiflo FQR and 



Ex. 18. The aidos AIS, BG and AC of u tr 
Bcribed circle at the poiuts I', Q and U. Show 
one-balf angle A are complementary. 

[Suo. Draw radii from the center to P and IE. Then I PQR 
IFOA, etc.] 

El. 19. Fi-om the point in whi 
angle meets the circumforeiieo, a ch 
of the angle. Show that this chord equals 
the other aide of the angle. 

Ex. 20. Given AB a diameter, AP= 
the radius, AD and PC iBngents; prove 
A CED equilateral. 

[St'G, Draw OC and CA; then in rt. 
A OOP, C^ = radma, ZP^GO", etc.] 

Ex. 21. Perpendiculars are drawn 
from the extremities of a diameter upon 
a tangent. Show that the points in which ihu perpendi 
the tangent are equidistant from the cente:. 




CONSTEUCTION PROBLEMS 

270. Postulates. As statod in Art. 49, a postulate In 
geometry is a coustrmitiou of a geometric figure admitted 
as possible. 

The postulates used in geometry, aru as follows (see 
Art. 50) ; 

1. Through any two points a straight line may he drawn. 

2. A straight line may J>e extended indefinitely, or it may 
be limited at any given point. 

3. A circMmference may he described about any given 
point as a center and with any given radius. 



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CONSTRUCTION PROBLEMS 147 

271. Tlie meaning of the postulates is that only two 
drawing instruments are to he used in making geometri- 
cal constructions; viz., the straight-edge niler and the 
compasses. 

With these two simplest drawing instruments it is 
ble to be able to construct as many geometrical 



272. Form of solution of a problem. The stiitoment of 
a problem and of its solution consists of certain distinct 
parts which it is important to keep iu miiid. These are 

1. The general enunciation. 

2. The particular enunciation. 

(1) Given, etc. 

(2) To construct, etc. (or some other construction 

phrase, as "to draw," "to bisect," etc.). 

3. The construction. 

4. The assertion. 

5. The proof of the assertion. 

6. The conclusion (indicated by Q. v.. p., quod erat 
faeiendum, "which was to be done"). 

7. The discussion of special or limiting cases, if such 
cases oceiir in the given problem, 

la the figures drawn in eonnection with probleins, tbe (lircii U)ies 
»fe drawn as hearij lines, the lines required fts light liiu-s, and the 
oasiliary lines ea dotted lines. 



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.ANE r.EO.MF/J'RY 



PROPOsri'io>; XXIII. Problem 

273, From a given point wilhout a give)i line to draw 
a perpendicular to the line. 



Given the line AB and the point 7* outsiilo AB. 

To construct a pRi-petidicular from thi^ point P to the 
line AB. 

Construction. With P as a center and with any oonve- 
iiient i-iidins, describe an arc intersecting A B in two points aa 
at C and D. Rost, 3, 

From Cand J> as centers and with convenient equal 
radii, greater than J CT>, describe arcs intersecting at Q. 

Poet. 3. 



[Assertion]. Then PE- is the 1. required. 

Proof. P is equidistant from the points C and T>. Constr. 

Also Q is equidistant from the points C and B. Constr, 

.-. PR 1 CD. Art. 113. 

(two joints, eadli eqiiklistant from the exlremifies of a thie, determine the 
X Mseetor of the line) . 

Q. E. F. 



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CONSTRUCTION PROBLEMS 



Proposition XXIV. Problem 

given line to erect a per- 



274. At a given point in 
pendicular to that line. 



Kg. 1 Fig. ,3 . 

Given the point P in the line A B 

To construct a perpendicular to the line A B at the point P. 

Method I. Construction. From V (Fig. 1) as a center 
with a convenient radius, describe an arc cutting off the 
equal segments PC and PB on the line AB. Post. 3. 

From C ond I> as centers and with equal radii, greater 
than PI), describe ares intersecting at li. I'oai. ?.. 

Draw PR. l-oBt.l. 

[Assertion], Then PR is the _L required. 

Proof. Let the pupil supply the proof. 

Method II. Construction. Take any point (Fig. 2) 
without the line AB, and with OP as a radius, describe a 
circuniferenee intersecting the line AJi a.t C. Post. 3. 

Draw CO, and produce CO to meet the cireumference 
at R. Draw KP. Posta. I and 2. 

[Assertion] , Then RP is the ± require<i . 

Proof. Let the pupil supply the proof. 

Discussion. When is Method II preferable 1^ 



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BOOK II. PI,AXR GF.OWK'r: 



Proposition XXV. I'koblem 
275. To biseci a given liHfi. 



Given the line AB. 

To bisect line AB. 

Construction. With A and B as centers and with equal 
radii, greater than i AB, describe ares intersecting in 
€ and t). Post. 3. 

Draw the line CD intersecting AB in F. Post. I. 

Then AB is bisected at the point F, 

Proof. Let t]ie pupil supply the proof. 



PkOPOSITION XXVI. PltOBLEJl 

276. 2'o hisfct a given arc. 




Given tlie arc AB. 
To bisect arc AB. 



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CONSTRUCTION PROBLEMS 151 

Construction. With A and B as centers, and with con- 
venient equal radii, describe arcs intersecting at G and Z>. 

Post. 3. 

Draw the line CT> intersecting the arc AB at F. Post. 1. 
Then are AB is bisected at the point F. 
Proof. Draw the chord AB. 

Then CD X chord AB at its middle point. (Why?) 

.-. CD bisects the chord AB, Art. 223. 

{the X hiseulor of a chord jiimscs Oirovgh the <:':nUr and Inserts tlie arcs 

siihteiKlal by the chord) . 



Q. E. F. 



Proposition XXVII. Probleji 
277. To Used <t oitrii aii'jh. 




Given angle A OB. 

To bisect angle A OB. 

Construction. With as a center and with aiij' conve- 
nient radiu.H, describe an arc intereeeting OA at C and 
OB at J). Post. 3. 

With C and I> as centers and with convenient equal 
radii, describe arcs intersecting at F. Post. 3. 

Draw OF. Pest. 1, 

Then lAOB is bisected by line OF. 

Proof. Let the pnpil suppjv the iit'ooE. 

)j.a.jf, 

fii- Construct ail I of iO°, 



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15a i^ook ii. i'laxe geometky 

Proposition XSVIII. Problejh 

278. At a ginen point in a glvrn .^Imi'jhl Vn-' fo c 
struct an mujlr equal to a given amjlc. 



Given thp Z A, iiod the point P iu the line liC. 

To construct at the point P an angle equal to A A, and 
having PCfoi- one of its sides. 

Conatnictiou, With .4, as a center ami with any eonve- 
nient radius, describe an arc meeting the sides o£ /.A 
at D and B. Post. 3. 

Draw the chord T^E. Post. i. 

With P as a center and with a radius equal to AT>. 
describe an arc cutting FC at F. Po&t. .t. 

From P as a center and with a radius equal to the chord 
DP, describe an arc intersecting the arc HF at G. Post. 3. 

Draw- FG. Post. i. 

Then Z GFF is the angle reqnired. 

Proof. Let the pupil draw the chord FG and complete 
the proof. 

Q. E. F. 

Ex. 1. On B given line oonBtmet a square. 

El, 2. Construct an angle of 30°. 

Kx, 3. H«nce eonBtruet an. angle of 15". 



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CONSTRUCTION PROBLEMS iM 

Pboposition XXIX. Peoblem 

279. TliTough a given point without a given straight 
line to draw a line parallel to a given line. 



Given any point P without the line A li. 
To construct a line through P II -4 7i. 

Construction, Through P draw any convenient line CT) 
meeting AB in C. Poat. i. 

At P in the line CD construct / JJPF equal to Z PCS. 
Art. 278. 
Then EF is the line required. 
Proof. Let the pupil snppiy Ihe proof. 

Ex. 1. Tlirough n given point dvaw a lino paraliel to a given line 
by conBti'utiting a parallelogram of which the given point is one 

Ea. 2. On a p;iven line as a diameter, oonKtniot a eirele. 

Ex. 3. Oonatruet an are of 00" having a radius of 1 In. 

Ei. 4, On the 9gnrB, p. 148, if the point Q be constructed below 
the Una JB, will ihe perpendicular required be likely t« be more 
accurately, or less accurately constructed ? 

Ex. 5. From a given point on a piven eireumference, how many 
equal chords can be drawn ? 

Ex. 6. Throuf;h a p;iveQ point within a given cirenniferenee, how 
many eqnal chords can bo drawn? 

Ex. 7. Froru a given point esternal to a given oircle, how many 
equal secants can be drawn ! 



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r,ANK GEOMETllY 



PiutrosiTiON XXX. Ph( 

280. Tn divide, n tjircn s/migM lim 
nHiiiber of cqind inni.-<. 



Given tiie line A B. 

To divide AB into a required unuiber (as Uiree) eqoal 
parts. 

Construction. From ^1 draw the line AT iiutkiiig a con- 
venient angle with AB. Post. i. 
Take AC any line of convenient length and apply it to 
AP a number of times equal to the number of parts into 
■which AB is to he divided. Post. 2. 
From E, the end of the measure when last applied to 
AP, draw EB. Post. 1. 
Through the other points of division on AP, viz., C 
and Ji, draw lines ll EB and meeting AB at F and G. 

Art, 270. 
Then AB is divided into the required number of parts 
at F and 6. 

Proof. AG = CD = DE. Constr. 

.-. AF=FG ^ GB, Art. 176. 

(if three or more parallels intercept cq^ial parts on one transucrsal, Ihey 
intercept e<pial partu on evenj traimversal). 



Ex. 1 . CoDstruet a right triangle whose legs are 1 in. and li in 

'Ef. %. Congtriiet a J ec tangle whose base 13 1 in. an<3 altitude 1 i: 



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COKSTEDCTION PROBLEMS IDa 

Proposition XXXI. Problem 

281. To construct a triangle, given two sides and the 
included angle. 



Given m and n two sides of a ti'iuugle aiul P the augle 
ineiiided by them. 

To construct the triangle. 

Construction. At the point A in the line Afl coustvuet 
I A equal to the given /.P. Art. 2TS. 

On the line AJi lay off AO equal to wi. Post, J. 

On the line AF lay olF .) I) equal to *i. Post, ••. 

Draw J)C. Post, 1. 

The A ^I>C' is the triangle required. 

1). E. F. 

Ex, I. Constniat a triangle in ivhk'h two of the sides are 1 iu. snil 
Itin,, nod the included angle is 13:)". 

Ex. 2. Construct an isoaeoles tiiun^lu, in wiiiHli Ihe liase shall bii 
H in, and the altitude 2 in, 

Ex, 3. Constniet llie complement of a given acute angle, 

Ex. 4i, Constniet the supplement of a given angle. 

Ex. 5. How is tiio figure on p, 154 i^onstrueteJ wilh Ihe fewest 
adjustments of tlie eompasses ? 

Ex.. 6. DniK a. line (segment) iiud mark olY lliree-iifllis of it 



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156 BOOK n'-. PLANE GKOMETRV 

Pboposition XSXII. Problem 
282. To consirud a triangle, givni tm angles and ihj 
included side. 




Given the A P and Q and the inr^luded side m. 

To construct the triangle. 

Consttuction. Take any line AI) anil on it mark AB 
equal to m. Post. 2. 

At A construct an angle equal to / P. Art, 278. 

At B construct an angle equal to Z Q. Avt, 278, 

Produce the sides of the A A and B to meet at €. 

Poflt. 3. 
Then A ACB is the triangle required, 

Q, E, F. 

Discussion. Is it possible to construct the triangle if the 
sum of the given angles is two right angles ? Why ? Is it 
possible if this sum is greater than two right angles ? Why? 

n wliiai two of the angles are 30° anii 
9 II iii, 

Ex. 2. Construct the complement of half a given angle. 
Ex.3, Coiistruot anaDgleo£120°; of I50°i o£ 135°. 
Ex.4. Trisect a given risht angle. 



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construction pkoblesis 

Proposition XXXHI. Problem 
;S. To construct a triangle, given the three sides. 




Given m, n tmdp, the throe sides of ;i triangle. 

To cottstruct the triiingle. 

Construction. Take the line AB equal to m. Post. 2. 

With A as a center and with a radius equal to n describe 
an are, and with B as a center and with a radius equal to 
p describe another arc. Post. 3. 

Let the two arcs intersect at the point C. 

Draw CA and GB. Post, \. 

Then A ABC is the triangle required. 

Discu^ion. Is it possible to construct the triangle if one 
of the sides is greater than the sum of the other two sides? 
Why? 

What bind of a figure is obtained if one aide equals the 
Bum of the other two sides 1 

Ex. 1. Cotisti'uet ac aiiijle of 22-i°. 

Ex. 2. Divide a given cireumlereiiee into four qundranta. 

Ex. 3. Construct the figure on page 82, using tin 
the tbcee altitudes aa & teat of the accuracy ot tbe woi 



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158 BOOK II. PLANE GKOMETltV 

PiiOPOSiTioN XXXIV. Problicm 

284. To construiit <i irkuujh:, given two sidr-f and an 
angle opposite one of them. 

Given »i and n two sides of a A and Z 1' opposite n. 
To conatmct the triangle. 

Construction. Several cases occur, according to the rela- 
tive size of the given sides and the size of the given angle. 

Case T, Wlim n > m {<uid I V U acute) . 






At the point A construct Z EA1> equal to Z P. Art. 27s. 
On AE take AB equal to ?h . Post. 2. 

With £ as a center and with a radius equal to h, describe 



an arc intersecting AD at C and C. 
Draw BO and BC. 



Post. 3. 
Post, 1. 

Two A, ABC and ABC, arc obtained, containing the 
sides m and n; but only one of them, A ABO, contains 
the ZP. 

.'. A ABO is the triangle required. 

Case 11. When n = m (and /.Pis acute). 
Make the same construction as in the preceding case. 
The arc drawn intersects the line AD in the points A 
and 0. 

Hence the isosceles A ABC is the triangle required, 



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CONSTKUCTION PE0BLEM8 



Case III. When n < m {and IP is acute). 




Make the eonstruetiou in the same way as in Case I. 

Two ;&, ABC and ABC, are obtained, each of whieli 
contains the sides m and n and an angle equal to / P oppo- 
site the side n . 

:. A ABC and ABC are the triangles required. 

DiscussioB. In Case I, if ZP is a right Z , let the pupil 
eoustruet the figure and show that there are two ^ answer- 
ing the given conditions. If ZP is an obtuse angle, let 
him construct the figure and show that there is hut one 
answer. 

In Case II, if Z P is right, or obtuse, what results are 
obtained ? 

In Case III, if ZP is acute and n = the ± from B to 
AT), how niaoy answei'S are there? also, if w < this ±, 
how many ? 

If / P is right, or obtuse, what result is obtained ? 

Ex. 1. Constniet a triangle in mhieh two of the sides are 1 iu, aud 
ii in., and the angle opposite ibe latter side is 45". 

Ek. 2. Construet a triangle in which two of tiie sides are IJ in. 
and 1 in., and the angle opposite the latter side la 45". 

Ex.3. Construct a triangle in which two of the sides are U in. 
snd i in., and the angle opposite the Utter side is 30°. 

Ex. 4. Construct the figure of page SO, using the coneurreuee of 
the throe blseutois as a test of (he atjcuraey of thu woik. 



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HOUK II. rLANE OEOME'J'ltY 



pKOfOSlTlOX XXXV, PifOllLKM 

285. To construct a -ihi rail dwj ram, i/Uru /iro sides and 
the inclmkd uiitjle. 






Given m and ii two sidos, arid P Uh' iiidiuli-d /. of a ITJ . 

To construct the parallelogram. 

Construction. Take line AB equal to m. Post. 2, 

At the point A eoiistrnet /.OAB equal to ZP. ah, ?7m. 

On the aide AG lay off AD equal to ". 

From D as a center and with a radius equal to ))i, Luui 
from B as a eenter with a radius equal to n, describe arcs 
intersecting in E. Post. 3. 

Draw ED and EB. Post. 1. 

Thun ABED is the parallelogram required. 

Proof. Let the pupil supply the proof. 



Proposition XXXVI. Problem 
286. To circumscribe a circle about a given triangle. 




Given the A AliC. 

To construct a circumscribed O about ABC 



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COXSTEUCTION PliOIiLEMS 161 

Construction. Ei'oct Js DFaud EG at tliu niidpoiiits of 
the sides AV auil AB, respectively. Art. 274. 

From 0, the poiiic of intersection of these A , witii a 
radius equal to OA, describe the O ABC. Post, -i, Art. iST. 

Then O ABC is the eii-cle required. 

Proof. Let the pupil supply the proof. 



Q. E. F. 



PKorosiTiox XXXVII. Problem 
287. To inscribe a cinie in a <jhcii Irianffle. 




Given the A ABC. 

To construct an Ui^^crihed circle in the triangle. 

Construction. Draw the Hue AD bisecting ^EAC, and 
Ci; bisecting- /lECA. Art. 277. 

From 0, the iutersection of AT) and CE, draw the liiie 
OP ± AC. Art, 273. 

From as a center with a radius OP, describe a O . 

P^-st. 3. 
Then circle ia the circle required. 

Proof. Let the pupil [supply tlic proof. 



Ex. 2. 1 
the tLree \ 



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162 book 11. plane geometry 

Peopositiox XXSVIII. Problem 

288. Throiujh a liiinii jicint on the circumfarence to draio 
a tanijmt to a cirvk. 




Given any pniot P on the (jircumference of the circle 0. 

To construct a tangent to tlie cirele at the point /'. 

Construction. Draw the radius OP. 

At the point P construct a line AB X OP. Art. 374. 

Then AB is tlie tangent required. 

Proof. Let the pupil supply the proof. 

289. An escribed circle is a circle 
tiuigpiit to one side of a triangle and 
to the other two sides produced. 
Thus the cirele ia an escribed 
circle of the A ABC. A center of an 
escribed circle, as 0, is called an ex- 
center Df tho triangle. 

Es. 1. Draw a triaagle nnd all of its cscribtJ cirtlRs. 
Ex. 2. Conatnict the figure on page 83, using tlie concurrence of 
the three medians as a Itst of the accuracy of the work. 




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COXSTKUCTION PSOELEMS lod 

Proposition XXXTX. Phoblem 

290. From a giw.n poini tvithout u circle to drain a iau' 
gent to the circle. 




Given F anj- point without the circk 0. 
To construct a line through F taiigeot to the circle 0. 
Construction. Draw the I'me PO. Post, : 

Bisect the line FO at ^f. Art. 37i 



From M as a center with Jff nri n 
cumfurcQcc iuterscctiiig the giveu 
and B. 



;, tkseribe a cir- 

iiftreiiee at A 

Post. 3. 



Draw FA and FB. 

Then PJ ami PIS are tlie tangents required. 
Proof. Z PAG is iLi>eribea in a semicircle. 
■■. IPAO is a ri^rht angle. 
.-. PA is tangent to tJie circle 0. 
In like manner PB is tangent to the circle 0. 



(Why!) 
(Why?) 



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164 BOOK II, I'LAXE r.Eo:in:TriV 

Proposition' XL. I'lionLKM 

291. Upon a given draUjM liiif to tlr^cribe a segmeni 
irhich shall conlitin a gh-eii- iiifcrihed angle. 




Given the ritraight line ^1 n and the Z P. 

To construct on the line ,17>' a segment of a circle such 
that any angle iuseribed hi tlie set^uieiit shall equal ^P. 

Construction. At the point B in the line AB constract 
ZABO equal to ZP. Art. 273. 

Constract DE the ± bisectoi- of line AH. Art. 274. 



From as a centev with OB as a radius, describe the 
circle AMB. Post. 3. 

Then AMB is the required segment. 
Proof, Let AQB be any Z inscribed in the segment 
AQB. 

Then /.AQB is measured by h are AFB. Ai-t. 258. 

But BG is tangent to the circle A2IB. (Why?) 

.". I ABC is measured by \ arc AFB. fWhy?) 

.-. /Q= lABG, or IP. (Why?) 

.*. any Z inscribed in segment A2IB ~ ZP. 

Q. E. F. . 



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constkuctiox problems 
Pkopositios XLI. Theorem. 



292. To find the common unit of measure of ti 
surable straight lines, ami hence the ratio of the lines. 



Given two lines AB luui C7>, of wliieli CT> is tlie shorter. 
To construct ii common unit of muaaiiro of J U luid CD, 
and iieiJCR obtain tlie rutb oJ: AB rmd CD. 

Construction. Apply CT> to AB as many times as 
possible; Post. 2, 

Say twice, with a remainder KB. 

Then apply KI^ to CD as many times as possible ; Poet, 2. 
Say three times, with a remainder LD. 
Apply Lll to KB as many times as possible; Post. 2. 
Say three times, with a remainder I'B. 
Apply PB to XD as many times as possible; Post. 3. 
Say it is contaiQed in LD exactly twice. 
Then FU is the common unit of meiiiuiu of -I /.' :i\v\ I 'H. 
Proof. LD=2PB. 

KD=^KP + PB^3LD + PB=1PB. Axs. G, 8. 
GD= CL + L7)^3KB + i/)=23P7i. 
^ B = .4 71 + /r /J = 2 CD + KB = r,3 I'B . 

A n _ r.r{ 7*/; „ :y.\ 

Hence -^ - ijj-^. = ^3" 



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SXEfiCISES IN CONSTRUCTION PROBLEMS 

293. Analysis of problems. The niothcxl of analysis 
(see Art. 196) is of especial value in tlii; solution of con- 
struction probleids. In general, to investigate the soiution 
of a problem by tliis method; 

Draw a fi<juri; in irJilch the vqi'irfil coiiKli-wtion is 
ansumed as made ; 

Draw auxiliary lines, ifnecflssriry ; 

Observe the rvlaiioHS hdwceii the purtg of this fujiire, in 
order to discover a known relaiion on ichiclt the required 
construction depends; 

Having discovered the required relation, construct another 
figure hy the direct «se of this relation. 

Es. Through a, given point within a oii'ole draw 
a chord which shall be bisected by the given point. 

Analysis. Let J be the given point witiiinthi? 
given circle 0, and let FQ be a chord bisected at 
the point A. A bisected chord suggests a line OA 
joining tbe point of bisection with the center 0, 
and that (Art. 224) OA ± PQ. 

Synthesis, or Dihrct Koluteon. Taking miotlipv IlKure contain- 
ing the data of the problem, connect the point A with the center of 
theeircle by the line OA. 

Through^ dniW a line 1 OA [Art,27+i, nnd meeting the circumfer- 
ence at tlie points I' and Q. I'Q is tho thurd rL-tinited. 

EXERCISES. CROUP SS 

CONSTRUCTION OF STRAIC.nT LINES. 
Ex. 1. Draw a line parallel to a given line, and tangent to a given 

[Sua. Suppose the req^uired line drawn; then the radius to the 
point of tangeney, if produced, is X given line, etc.] 

Ex. 2. llraw a Hub porpendicular to a given line, and tangent to 
& given circle. 




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EXERCISES. PEOBLEIIS 167 

Ei, 3. Prom two points In the eireumturenee of a eirele, draw 
two equal and parallel chords. 

Ei. 4. Through n giren point draw a line which ehall make a 
given angle with a given line. 

Ei. 5. Through a giveu point draw n line which shall make equal 
angles with the sides of a given angle. 

[Sua. The bisector of the s-Lven I will be ± the required Una, etc.] 

Ejt. 6. Through a given point between two given pai'allel lines, 
draw a line of given length with its extremities in the 
two parallel lines. /'^^'""\ 

Ex. 7, Through a given point J within a eirele, IA/ "•, \ 

draw a chord equal to a given line. V \ \ \ 

Ex. 8. From a given point in the oircumferenco V j_.^ <- 'J 

ot aoircle, draw a chord at a given distance from ^v.-^^!,-^ 
the center. 

Ex. 9. Througli a given point on the circumference of a circle, 
draw a chord which shall be bisected by another given chord. 

[SuG. Draw the radius to the given point and on it as a diameter 
describe a circle, etc. When is the solution impossible ?] 

294. Construction of points and of loci. In construct- 
ing a point to meet certain glvec conditions, it is often 
helpful to c<nisirw:i the locus of a point iinsu-erhig one of the 
given conditions and observe in ivhat point or points it meets 
a given line, or meets anotJier locus ansirei-ing another 
given condition. 

EXERCISES. CROUP 13 

COXSTBUCTIOX OP POIVl':: AXD LOCI 
Ex.1. Find a point Pin a given line Jii ^ p 

equidistant from two given points C and 11 '\^ 

[Suo. Construct the locus of all points ^ 

equidistant from C and D and observe where A~=^ \;__„^-_=^ 

it intersects the given line AB.] ' 

Ex. 2. Find a point 1' in a given circumfuicuce, which is equi- 
distant from two givi'n iioiritB, (.' and D. 



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1 givfu line, which i> 



Ex. 4. Find a point 
d, from a giren point. 

Ex. 5. Find n pnint whioli is at tl given ilistanee, a, from a givfn 
point A, riiiii at another distance, h, from anothiT givon i)oiiit B. 
Discuss llio limitations of tliis problem. 

Ex. 6. Find a point equidititant froia two givnn pointi^, mid at a 
given distance from a given straight line. 

[Srr.. Draw the locus of all points equidistant fi'ora the two given 
points, and also the locus of all poluta at the given distance from tho 
given straight line, etc.] 

Ex. 7. Find a point equidistant from two given points, and at a 
given distance from another given point. 

Ex. 8. Find a point equidistant from two jriven points, and also 
from two given intersecting lines. 

On the other hand, the determiaatioa of certain loci is 
equivalent to the conslrnciloii of all points which snti.tfy one 
or more given conditions. 

Ex. 9. Find the loeua of the center of a aivole, which touches a 
given line at a given point. 

[Sro. Construct a number of circles touching the given lino at 
the given point and observe the relation of their centers.] 

Ex. 10. Find the locus of the center of a ciraumterenoe with t 
given radius, r, which passes through a given fixed point. 

Ex, II. Find the locus of the center of a circle, touching two giver 
intersecting lines, 

Ex.12, Find tiie iocusof the c 
parallel linos, 

Ex. 13, Find the locus of the ( 
which toucliBS a given straight line 

Ex. 14, Find the locus of tiic t 
which touchcii a given circle, 



ir of a circle of given radin: 
■c of a eirale of given radiu 



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EXERCISER. EROBLEJIS 

EXERCISES. CROUP 24 

COXSTRIXTIOX OF KECTII.INEAK riGL^RE 

Ex. 1. An eijuilateral triangle, given the pecimeter. 

Ex. 2. An eqiiilaloral triangle, given the Rltituda. 

Ex. 3. An isosceles triangle, given tho bnse ami .iltiti 

Ex. 4. All isoseeles triangle, given the ba.=o mul nri i 



?cles 



ri:mj;'l., 



Ex. 6. A right tviangle, given n. leg and iho acute angle adjacent. 
Ek. 7. A right triangle, given li Ic},' and the acute angle opposite. 
Ei. 8. A right triangle, given tlio hj'potenuse and an acute angle. 
Ex. 9. A right triangle, given the hypotenuse and a leg. 

given the altitude and the sides including the 



Ex. 10, A 
vertical angle. 

[SUQ. Through tlie foot of the altitude d 
of indefinite length, etc.] 



p 1 altitude and 



Ex, 11, 



igle, g 



iiJes and the 



ilUtudf? 



ipon 



e of 



Ex. 12. A square, given the diagonal. 

Ex. 13, .4. rliomljHs, given the two diagonals. 

Ex. 14. A I'homljus, given one angle and one diagonal. 

Ex. 15. A parallelogram, given two adjaoent sides and an altitude. 



Ex. 16. A parallL'lograr 


.1, giv. 


;n a side, the aUltude upon 


that side 


and an angle. 








Ex, 17. A paralUdit-rt 


im, yi 


ven the diagonals and 


an angle 


included by tlK^m, 








Ex, 18, A quadrilalerul 


, b'ive 


I, th« sides and one angle. 





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170 



BOOK n. PLANE GEOMETRY 



295. Use of auxiliary lines ia constructing rectilinear 
figures. In constructing polygons, auxiliary lines are fre- 
(iuentlj' of service. Tinis it is often of especial value to 
c<mstrtici, first, eitlier the inscribed or the eircumseribed 
circle, and afterward tJie required triangle or quadrilateral. 



Construction. Drawaeirolo with radius 
equal to r. Draw a tangent at any point A. 
On this tangent mark oft All and AC eiicli 
equal i b. From B and draw tangents iJl'' 
and CF to tlia circle. TUen BCF is t!io 
required trianglo. 




EXERCISES. CROU^ SB 

CONSTHUCTIOXS. AUXILIARY LINES 

Coiistruet 

Ex. 1. An isosceles triangle, givon the base and the radius of the 
cireurasoribed eiraie. 



Ex. 3. A right triangle, given (he radius of the iiircumscribed 
eirele and an acute angle. 

Ex. 4. A right triangle, given the radius of tlje Inscribed oirele 
and an acute angle. 

[SUQ. Draw the inscribed circle and at its center constrnet au 
anele equal to the supplement of the given angle.} 

[. 6. A triangle, given the base, flio altitude and tho vertex 



-ingle. 



a construct n segment which sliall contain 



[Boo. On the given 
the given vertes angle. See Art. 291. J 

Ejc. 6- A triangle, given the baso, the mL'dian to the base, and the 
vertex angle. 



Ex. 7. A triangle, given one 
of the inscribed circle. tSue Kx. 



angle, 



nd the radius 



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EXERCISES. PROELEMH 17X 

II adjacent angle, and the 

The use of mtxiliary straight lines may be illustrated as 
follows : 

Ex, 9. Constmct a triangl6, ei^'^i the perimeter and two angles. 

Analysis, Suppose the required triangle JBC already eonstriioted 
and let AAISC and JCifbe the given 

angles. Produce £C to D and E, ^ 

making DB = AB and CE =^ AC. -■-'/'^V- 

Tban DE= given perimeter. Also --'"'/ \ '""""- 

Similarly IACB=21E. Hence D B C £ 

CoKSTBUCTiON, Take D£tlie given perimeter; at 75 ctingtruot an 
angle = X of 0"6 given angle! "t E eouatriict an angle = % ot the 
other given angle. Produce the sides of these angles to meet e,t A. 
Construct ADAB = ID and ICAE = IE. Then A ABC is tbo 
required triangle, ete. 
Construct 

Ex. 10. An iaoscelea triangle, given the perimeter Kud the 
altitude. 

[Sue, Bisect the perimeter and eonstruet the altitude ± to it at 
itsmidpoint-1 

Ex. 11. An isosceles triangle, given the perimuter aud the vertoi 
angle. 

[SUQ, If the vertL'S Z is known, tiie base i may be obtained.] 

Ex, 12. Arighttriangle,eiyonnnaciitH ^ 

angle and the sum of the legs. •' t"""--^ 

[SOG. Given AB the sum oC the legs, ,-■'' ^''~~--^.^^^^ 

eonstruet iiA=ii)°, etc.] _^ j) - ■- ^ 

Ex. 13, A riglit triangle, given an ueuto angle and the difference 
ot the legs. 

Ex. 14. A right triangle, given the hypotenuse and the sum of 
the legs. 

Ex. 15. A riiilit triangle, given au a«utu angle iind Ihs auoi o£ the 
liyjotenuse and unu kjj. 



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172 liOOK II. rL.VNR GEOJIT-.TRY 

Ex. 16. A triangle, given an s.ng\-:, a siile nM thn sum of ti 
Other tTTo sides. 

Ex. 17. A ti'inngle, given an angle A, the sum o£ tlie sides j 
and BC and tlie altitude upon AB. 

296. Reduction of problems. Iti many caao.s a prolxhn 
may be solved hy reducing the proilcm to a problem alreaii 
soloed. (Tliis is a special kind of analysis.) 

Kx. Construct a parallelogram, giycn the diagonals and one side. 

Analysis: Suppose the £ZJ ABCF to bo Ihfl required /H? alread 
constructed. Let AF be the given side, If the diagonals ni 
given, halt each diagonal ia given (Art, 

161). Hence in the A AOF the three j? 

sides are given. Heneo the recinired / "^-.^ _,-■ 

problem reduces to the problem of con- / -''n'- 

structing a triangle whose three sides •..--'''' 
.are given (Art. HS3), Henee 

Construction. Let the pupil supply tho di 



EXERCISES. CROUP 26 

Rl'IDrCTION OF CONSTRUCTION PROBLEJIS 
Construct 

Ex. 1. A right triangle, given the aitituily upon tliy iiypotenusE 
and ttie median upon the sniue. 

Ex. 2. A rectangle, given the perimeter and a diagonal (see 

Ex. 3. A rectangle, given the perimeter and an angle made by 
the diagonals. 

Ex. 4, A triangle, given the Ihreo angles and tlie railius of th» 
circumscribed circle. 

[SuG. The sides of the A are the chords of 11 
segments of the O eontaiuing the given i .] 

Ex. 5, A triangle, given two aides and the median / / 

to the third side. j/' 



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EXEKCISES. PROBLEMS li6 

Ex. 6. Atrianftle, Riven the thi'ee medians. 

[Suo. Beauee this to the pveeeding Ex. See Art. 187.1 

Ex, 7. An isosceles trapezoid, giveu the buses and ati angle. 
Bs. 8. An. iaoaceies trapezoid, giveu the biiaes and a diagonal. 
Es. 9. A trapezoid, given tlie four sides. 
Ex. 10. A trapezoid, given the bases and tiio two diagonals. 
[Src. Reduoo to Art. 2S3 by producing, tlie lower base a distance 
equal to the upper base, etc.] 

297. Construction, of circles. The consfcraetion of a 
required circle is freciuently a good illustration of tlie 
preceding method of reducing one eonsfcruction problem to 
another. For the construction of a circle frequently re- 
duces to tlie prohlem of fiitdhuj a poitit {the center of the 
circle) which answers giren eonditioiis. (See Art. 294.) 

Ex. Construct a circle which shall touch two given intersecting 
lines and have ita center in another given lino. 

This problem is equivalent to the problem of finding a point which 
shall be in a given line and be equidistant from two other given 
lines. (See Ex. 3, p. lOS.) In soma cases, however, the eonstrno- 
tion of a requirud t^irolii iiiiift be made by an independent method. 

EXERCISES. CROUP 17 

COSSTEUCTIOX OP CIRCLES 



Construct a circle with given 


, radius, r, 




Ex. 1. -Which passes thr. 
line. 


ough a given pfiint and tf 


Kiches a given 


Ex. 2. Which has its ce' 


uler in a given line uud t, 


luehes another 


given line. 






Ex. 3. Which passes thri 


)ugh two given points. 




CocHtruet ft cirelo 






Ex. 4. Which touches tw 


•0 givou parallel lines and 


passes through 


a given point. 







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Ex. 5. "Wiiii.-li parses llinin^'li two gi^-Lii 
on a given line. 

Ex, 6. Whiuh toucho3 three given lines, 
two of which are paraHcl. 

Ei. 7. Which passos through a givon 
point A and touches a giren lice liC at a 
given point J>'. 

[SOG, Draw An tmd at li coiiatriLfl ;i 1 
to fiC] 

Ex. 8. TV'hicli toiithes a givenlins and al: 
touches s, given circle u 
/ \ a given point ^. 



..o 



Ex. 9. Whieh touchefi .i given line AJl 
t a given point A and touches a given 



EXERCISES. CROUP 28 

PROBLEMS SOLVED BY VARIOUS METHODS 

Kx. 1. Through a given point draw a line which shall cut two 
given intersecting lines so as to form an isosneles triangle. 

Ex. 2. Construct an isosceles ttiimgle, given the altitude and 
one leg. 

Ex. 3. In a given circumference find a point equidistant from two 
given intersecting lines. 

Ex. 4. Draw a circle whicli shall toui^h two given inlerseoting 
lines, one of thein at a given point. 

Ex, 5. Draw a lino which shall lie terminateJ by the sides of a given 
angle, shall equal a given line, and be parallel to another given line. 






1 ndjac 



mgle, 



Ex. 7. Find a point in a, given circumference at a given distanea 
from Si given point. 



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MISCELLANEOUS EXERCISES. PROBLEMS 175 

Ex. S. Construct a parallelogcam, given a siJo, an angle, and a 
diagoniil  

Ex. 9. Through a given point within an angle, draw a straight line 
terminated t>y the sides of the angle and bisected by the given point, 

[SuG. Draw a line from the verte-i: of the angle to the given point 
and produce it its own length through the point.] 

Ex. 10. Construct a triangle, piven the vertt'i angle and the 
segments of fhe base made by the altitude. 
[SuQ. Use Art. 291,] 



n ciidiiis which shall touch a 



Ex. 11. Construet an 


isoscclf 


ertcxand the base. 




Ex. 12. Drawacircli 


5 with si 


irele at a given point. 




Ex 13. Construct a 


right tri; 


Iti tilde upon the hypoto: 


QUSO. 



■, given the hypotenuse and the 



Ex. 14. Construct a triangle, given the base and the altitudea 
npon the other two sides. 

[SuQ. Construct a semieircle on the given base as a diameter.] 

Ex. 15. Find a point in one side of a triangle equidistant from 
the other two sides. 

nd the angles at 



Ex, 17. Construct a rhombus, given an angle and a diiigonal. 



Ex, 20. In a given circle draw a chord equal to 
parallel to another given line. 

[SuG. Find Ibe distance of the given chord fro: 
couattiieting a right triangle of which the hypotouuse 
given,] 



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1(G 1iOOK II, PI.AXE GEOMETKY 

Ex. 21, Consfrupt a ti'in!if;Io, giv'ii nil aiL.ijii;, thu bisei'tor of that 
angle, and tbe aititude from iiuotlitir vertox. 

Es. 22. Find the loeus oE the points (jE ciontact oE tangL'tils dfawa 
from a given point 1o ii hltIl's of circlea kaviii!!; a given i^tuter. 
[Sm, Uao ArU. il'JD mid 2til.] 

Ei. 23. Given a line ,17; ami Uvo points. -G 

C and D, on the same side of JJl. Find a 
point P in ^J! such that IAPC= IJU'D. 

[Suo. Draw a 1 from C to J7! ajid pro- j^ +^ ^ 

dueeit its own length, ecc] 

Ex. 24. Given a line -17i and two points r and }> on the samo side 
of AB; find a point P in J7( suth that CP + I'D sliall ho a minimum. 

Ex. 25. Draw a 
tangent to two given eireh 

Ex. 26. Dra' 
tangent to two given circles. 




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Book III 
PROPORTION. SIMILAR POLYGONS 

THEORY OF PROPORTION 

298. Ratio has been defined, and its use briefly indi- 
cated in Arts. 245, 246. 

299. A proportion is an eKpressioTi of the eriiiality of 
two or more equal ratios. As, 



This reads, "the ratio of d to J equals the ratio of c to 
d," or, "a is to i> as c is to d." 

800. The terms of a proportion are the four quantities 
used in the proportion. In a proportion 

the antecedents are tlie first and third terms; 
the consequents are tlie second and fourth terms; 
the extremes are the first and last terms; 
the means are the second and tJiird terras. 
A fourth proportioaal is the last term of a proportion 
(provided the means are not equal). 

Thus, in « ; fe = c : d, dis a. fourth proportional. 

301. A continued proportion is a proportion in which 
each consequent and the next antecedent are the same. 

Thus, a : & = & :c = c : d = d : ^ is a continued proportion. 

A mean proportional is the middle term in a continued 
proportion containing but two ratios. 
h (IT") 



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1(8 ISOOK III. PLANE GEOMETliY 

A third proportional is the last term In .i coiitiimod pro- 
portion containing but two ratios. 

Thus, m n : h^h : c, b i& a. mean pi'oportioual, and c is a 
third proportional. 

Phoi'OSitios I , Theorem 

302. In any proporiion, the prndiict of the extremes is 
equal to tlie pvodwl of tlie wniiis. 
Given tlie proportion a : b = r : d. 
To prove ad = be. 

Proof. f^ = 2- "^i"- 

Multiply each memher by hd. 

ad — he. Ax. i. 



Proposition II. Theorem 

803. TJie mean proportional iHween two quantUie; 
equal to ike square root of I heir product. 
Given tlie jtroportion a :l) = h : 
To prove b = y( 

Proof. a:l>^h : 

:. h" = ac, 

(in any projfortion, the i>rodiict of th- 



.-. b=v;^ 



Hyp. 

Act. 303. 

equals the product of the 



Q. E. B. 

Ex, 1. Find the fourtli proportional to 2, 3 and C ; also to 3, i, ^. 
Ex. 2. Find tlie mean proportional between 3 and 6. 
Ex. S, Find the third proportioual to 3 and 5. 



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THEORY OF PROPORTION 179 

Proposition III. Theoresi 

304. If the product of two quantities is equal to the prod' 
net of two other quantities, one pair may be made the ex- 
tremes and the other pair the means of a proportion. 



Given ad = hG. 




To prove 


a , b^c: (I. 


Proof. 


ml- Ik. 


Divide eacli memlici 


 l.y M. 


Then 


b' d 


Or 


o :S-C ; <!. 



306. Cor. 1. If the (uitecedeuts of a proportion are 
equal, the consequents are equal. 
Thus, if a:x = a\y, then x = y. 
Let the pnpil supply the proof. 

306. Cor. 2. If three terms of one proportion are 
equal to the correspondinrj three terms of another proportion, 
the fourth terms of the two proportions are equal. 

Thus, if II : h = c : x, and a : li = c -. y, then x = y. 

Let the pupil supply the proof. 

Ex. 1. 'Write <ih = i>q as a proportion in as many different ways as 
Ei 2. Write a:(j:+l) = (i na a proiwtiou ; ulao j;- = 15. 



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Proposition IV. Tueokem 

807. If four quantities are in proportion, they are in 
proportion 6j^ alternation ; that is, the firnt term in to the 
third as the second is to tlie fourth. 

Given the proportion a •.b = c -. d. 

To prove a . c=h -. d. 

Proof. a : h^c : d. Hyp. 

.-. (id = bc, Art. 302. 

(in anu pronortimi, ike product of the extrnauis equals the iiroduct of 

Writing a and d as the extremes, and c and h as the 
means of a proportion, 

a : C = l) : d. Art. 304. 

(*/ the product iif tieo qnaiitities is eqiiiil io ih^ prodni-t of iiro nlher 



Proposition T . Theorbm 

308. If four quantities are i» proportion, they are in 
proporHow 6j/ inversion ; that is, the second term is to (he 
first «s the fourth is to the third. 

Given the proportion a ■.h = c -, d. 

To prove Ji : a^—d -. c. 

Proof. a : h — C: d. Hyp, 

.-. ad^bc. (Why?) 

Writing 6 and c as the extremes, and a and d as the 
means, 

I : a^d : e. (Whylj 

Q. E, ». 

Ex, TranB£ormi:a = i):o bo that j: shall be the last term. 



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TlfROliY OF PROPOKTION 181 

Peoposition VL Theorem 

309. If four quanfitips are in proportion, they are in 
proportion by compositioa ; iktif ix, the. snm of the first two 
terms is to the second term ws the sum of the last two terms 
is to the last-term. 

Given the proportion a : h — c -. d. 

To prove a -{- h -. h^c-V d: d. 

Proof. 7= T ^?P- 

h a 

Add I to oach meiuber of the equality. As. 2. 

h a 

^ b d 

That is a + h -.h^c + d: d. 

Let tlie pnpil show also that a -r h : n = c-{- d : c. 



Froporitios- VIT. Theorem 

310. If four quindiUcs are in proportion, they are in 
proportion ?!(/ division ; that is, the difference of the first 
two is to the second as the difference of the last two is to 
the last. 

Given the proportion a -. b = c : d. 

To prove a — b:l> = c — d:d. 

Proof. ^ =^- Hyp. 



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18'J HOOK iTt. i=LA>rE nroMi 

Subtratit 1 £min vM-h mi>,m)Mv of I he • 

1.1 d 

a — h '■ — >! 

That is fl-?i : i = c — d : (L 



Q. E. D. 



Let the pui>il hIkiw also that a^b • a—c — d  c. 



PRorosrnoN VTII. Theorem 
Sll. If Jovr quaniities are in proportion, they are in 
proportion hy composition and division; that is, the sum of 
the first two is to their difference as the sum of the last two 
is to their difference. 

Given the propoi'tioii a -. b = c : d. 
To prove a-\- b -, a — !> = <■ -\- d -. r, — d. 
Proof. *' ; /' = '■ : 'I. Hyp. 
^'-^h_c^d_ 



By cooi|msifii! 



d 



Dividing the coiTespoiiding men 
equalities, 

a+b^ c+d 
a — h e — d' 

That is a-^ h : a — & = c + d -. 



Ex. 1. WhHt do«e the proi-iortion 12:^ = 8:2 beeoi 
tionT also ty divisiout 

Kl. 2. Whatdoea 2.c — 5:5 = 3^: — 7:7 becomoby t 



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THEOBY OF PROPORTION lod 

Proposition IX. Theorem 

S12. In a series of equal ratios, fhe svm of all (he ante- 
cedents is to the sum of all the consequents as any one ante- 
cedeni'is to its consequent. 

Given a -.l^c : d = e if^g : h. 

To prove a-\- c-\- e-}- g -. J> + d+f+ h = a -. h. 

Proof. Denote each of the eqnul ratios by r. 

Then l^r, :. a^br. 

Similarly c — dr, e=fr, g — hr. Ax. i. 

Hence a + c + c + q={b -\- <l -i- f + h) r. Ax. 2. 

Dividing by b i- d +f+ h, ^qrf+y:^ = *" ^ ^ ' ^'^- ^■ 
That is a + c-<re + g -.h + d + f+h^^aa. 



Proposition X. Theorem 

313. The protiucts of the correspovllut/ /ci-Jiis of two or 
more proportions arc in proportion . 

Given a : I'^c : d, e : f^g -. h, ami j : A-=i : m. 
To prove aej ; hfk^cgl : dhn. 

Proof. 7= -,. 7= f , T= -■ Hyp. 

h d f h k. i)i 

Multiplying together the uorvetipoiiilijig terms of these 

equalities, 

Ai. 4. 

Q. E. B. 



aej _ cgl 
t)fk dh7n 
That is aej : bfk^cgl : dhm. 



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54 EOOK III. 1>T,ANF. CJEOMETIiV 

Proposition XL Theorem 

314. Lilie 'powers or like roots of ilie terms of a proper- 
on are in proportion. 

Given the proportion » -. h^c -. d. 

To prove o" : V ^ c" ■. i!", and a" : h" = v" -. d". 

Proof, 7= V Byp, 

Raiding both loembers ui the h"' power, 



That is a" -.h" =c" :(!". 

In like manner ci" -. ?i" = c" : d". 



Proposition SII. Theorem 

815. Equimultiples of two quantities have the same 
ratio as the quantities themselves. 

Given the two c|naiitities a and 6. 

To prove ■ma -. viJi = a -. b. 

Proof. 7- J- Went. 

Multiply eaeh term of flie first fraelion by m. 
ina_ a 
■'■ ml>~ h 

That is ma : mh = a -. h. 

__^_________ Q. E. B. 

Ex. 1. Transform ji : q = x : j in all possible ways by the use of the 
properties of a proportion. 

Ex. 2. TraQB£ovm 7:3 = 28:12 by eompoaitiou aiiJ diTiBion. 
Ex.3. Also2a; + 5;2i-5-a^+l:j;' — 1. 



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THEOBY OF PEOPORTION 18.) 

816. Note. In the theory of proportion as just p^eaented, the 
CjuaQtities used are assumed to be ooinmenaurablo, but the saiue 
theoreniB may also be proved for proportiona wlioae termB ara iucom- 
menaurable by use o£ the method of limita. For each incommenan- 
rable ratio may be made the limit of a eorrespondiug eommenaurable 
ratio; then, by ahowing that the variable commensurable ratios are 
equal, it may be proved that the limiting incommensurable ratios 

It is also to be noted, that, in the above theorems, the terms of a 
ratio must be of ttie same kind of quantity; that is, both be linea, or 
both besurfacus, etc. Hence, in orderthat aproportion be treated by 
alternation, for instance, all four of the terms must be of tlio same 



Ex. 1 . Find the fourth proportional to a, 2<i, 3 ic. 

Ex. 2. Find the third proportional to rt -;- '< and a — h. 

Ex. 3. Find the value of x in the proportion, 4 : 5 = a ; 15. 

Bx. 4. Findft shortmethodof determining whether a given propor- 
tion is true or cot. Use this method to determine whether the follow- 
ing proportions are true: (1) 4 r 6 = 3 : 9. (2) 5a : 2o = 15 : 6. 



Ks. 6. Construct exactly the figure of page 77 with the fewest pos" 
sible adjuatmenta o£ the compasses. 



Ex. 7. Draw an obtuse triangle and 
uaing the eoncuiTeuce of the altitudes as a t 

Ex. 8. Arrange nine points in a plane 
greatest number o£ straight liues may pasa i: 
pass through three points and only three. 



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EOOK III. TLAXR flFllMETP.Y 



PRorosiTiox Xlll. Tin:oBKM 

317. A line panilhl l'< ime, -ihlc of a triangle and meet 
ing the other two skUs, dlfides these sides proportionally. 




Given the tviaiiKle ABC ami the Urn- DK \\ base liC and 
intersectintr the sides AH and AC in ilie points 1) and J-], 
respectively. 

To prove BB -. AD^EG : AE. 

Case I. When T>B mid AD (Fig. 1) are commensxirabje . 

Proof. Talie iiny cornnion unit ..f niKisiiro of I)B and 
AD, as AS', and let" it be .'Oiitainu.l iu /)/.' a cerlain lunnWv 
of times, as n times, and in AP, m \\mv^. 






DP, 



Tbroiigli the points of division of DP. and AB drai^ 
lines II BC. 

These lines wiil divide EC into i\ . and AE into m parts. 
all equal. 



EC_n 
•'• AE in 

DB^EC 
 AD AE 



Art. 17G. 
(Why ?) 



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PROPORTIONAL LINKW 187 

Case II. When DB and AD (Fig. 3} are incommenKurnlile. 

Let the line AB be divided into any number of equal 
parts and let one of those parts be applied to DB. It will 
be contaiued in DB a certain onraber of times, with a line 
PB, less than the unit of measure, as a remainder. 

Draw PQ \\ BG. 

Then DP and AD are commensurable. Constr. 

 AD AE '^^' 

If now the unit of nieaKure be indefinitely diminished, 
the line PB, which is less than the unit of measure, will be 
indefinitely diminished. 

Hence DP ^ DB, and EQ ^ EC us alimiL Art.'isi, 

DP 
- AD 

EQ 

) ^^„.. .. ........ ^ 

DP ^, . -,-, EQ 

— ^thevanablej^a 



But the variable 



;. the limit ^-^,= tho limit f|- Avt. ^.w. 

AD AE 

d. E, D. 

318. Coit. 1. By eoniposition. Ail. ".09. 

DP + AD : AD^EC -{■ AE : A E. 

Or AB : AD = AG i AE. 

In like manner AB -. DB = AC -. EG, 
or, in general language, if a line parallel to the base cut the 
sides of a triangle, a side is to a segment of that side as the 
other side is to ilie corresponding segment of the second side. 

319. Cor. 2. Using Fig. 2, 

m^AL I!^=A^ PTJ^DP 

QO AQ' ^^^^ EQ AQ " QG E</ 
Henee. if two lines are cut by a number of parallels, the 
corresponding segments are proportional. 



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188 IKKlK III. PLAXE GROMETlii- 

Proposition XIV. Theohi-.m (Convehse op Vmn'. XlII) 

820. If a stffiinht line diviiUn iiro si'h-n of a triangle 
proportionally, it is iMraUel to the third side. 



Given the A ABQ and the linu DF intersecting AB and 
AG BO thut AB : AB^AG : AF. 

To prove DF]] BG. 

Proof. Through 7> draw the line BK \\ BG and meeting 
the side -iC in K. 

Then AB : AD = AC : AK, Art. sis. 

(if a line II bam i^v,t lite sides of a a, " aide is to a segment of ilial side aa 
tii^ other side is to the corresponiling segment oj the second side). 

But AB : AD-^^AG : AF. Hyp. 

.*. AF^AK, Art, 306. 

[if three terms of oni! ^roporthm are cijiinl to the enrrespouding three 
terms of another prn/wtiiHi, Ihr.fiiiirlli terms of She tiro ' 
j,roi>orUv)is arcapial). 

Ilenee the point K falls on F, and the line 1)K coin- 
cides with the line DF. Art. 6C. 
But the line DK 11 BC. Constr. 
,-. BF !! BG, 
(for DF coincides Kith DK, ivkicii is || BC). 

Q- E. D. 

Ex. 1. In the figure of Prop. XIII, if AD = 6, DB = 4, and ^£=9, 
find EC. 

Ex. 2. Also, if AD=12, Dii = S, aad AC^la, find AE aaii EC. 



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EXERCISES. CROUP IB, 

REVIEW ESERCISKS 



Make a list o£ the propertii': 
Ex. 1. One straight lin 
be related to the line). 

Ex. 2. Two straight lines that meet, or intei 
With such angles as may be formed by the given, li 

Ex. 3. Two or more para,llel lit 
and angles formed). 

Es. 4. Eight angles. 

Ex.5. Complementary angles. 

Ex. 6. Supplementary angles. 

Ex, 7. A single triangle (in 
triangle, aa altitudes, medians, et 

Ex. 8. A right triangle. 

Ex, 9. An isosceles triangle. 

Ex. 10. Two triangles. 

Ex, 11, Two right triangles, 

Ex. 12, A rjuadrilatersl. 



with RHph points as may 



nnneotion with lines within the 

Ex. 13. A trapezoid. 

Ex. 14. An isosceles tvapeioid, 

Ex. 15. A parallelogram. 

Ex. 16. A rhombus. 

Ex. 17. A polygon. 



Ex. 18. 


A single 


related poi' 


flts). 


Ex, 19. 


Arcs of 


angles) . 




Ex, 20. 


Chords 1 


Ex. 21. 


Central a 


Ex. 22. 


Tangents 


Ex. 23. 


Seeants 1 


Ex. 24, 


Two ciR' 



niferei 



(ir 



rotated lines ai 
^-ith related lie 



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190 HOOK in. PLANE OEOJIETRY 

SIMILAH POLYGOWS 

321. Bef. Similar polygous are polygons having their 
homolosroiis Liiigliis et^iial and their homologous sides pro- 
portiooal. 




Thus, if tliR figures AB€7)E and A'B'C'B'E' are simikj, 
the angles A, IS, 0, etc. , mnst equal the angles A', B', C, etc., 
respectively; also ^B : A'B'^BC : B'C'^ GJ) -. OB', etc. 

Hence it is cOiiMtantly to be borne in iniud that simi- 
larity in shape or form of rectilinear figures involves two 
distinct properties: 

1, The liojiiologoiis angles are equal. 

2. The homologous sides are proportional. 

It should also be clearly realized that one of these 
properties may be true of two figures, and not the other. 

Thus, in the rectangle A and the rhomboid B, the cor- 
responding sides are proportional but the corrt 
angles are not equal. 



!i 



Also, in the reetangle G and the square J\ the corres- 
ponding angles are equal but the corresponding sides are 
not proportional. 

However, it will be found that, in the case of triangles, it 
one of the two properties is true, the other must be true also, 

322. Def. The ratio of similitude in two similar figures 
is the ratio of any two homologous sides in those figures. 



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SIMILAR POLYGOKS 



Pboposition XV. Theohem 

323. If two triangles are mutually equiangular, they 
are similar. 





Given t!ip & ABO and A'B'C witli lA^ lA', IB = 
AB', and / 6' = Z t". 

To prove tlie &^ ABC and A'B'C similar. 

Proof. Tlic given & are niutuaEly equiangular. Hyp. 

Hence it only remains to prove that their homologous 
sides are proportional. Art. 321, 

Place the A ^.'S'^upon the A ABO, so that I A' shall 
coincide with its equal, the ZA, and B'C take the posi- 
tion FH. 

Then lAFR = IB. Hyp. 

.-. FHWBC. (Why?) 

.-. AB : AF^AC : AR. Art. 317. 

Or AB : A'B' = AG : A'C. Ax. 8. 

In like manner, by placing the A A'B'C upon the A 
ABC so that the Z S' shall coincide with its equal, the ZB, 
it may be proved that AB -. A'B'^BC : B'C. 

Hence the A J,£Cand A'B'C are similar. Art. 331. 

324. Cob. 1. If two triangles have (wo angles of &ne equal 
to two angles of the other, the triangles are similar; also. 

If two right triangles have an acute anijle of one equal to 
an acute angle of the other, the triangles are similar. 

325. Cor. 2. If two triangles are each similar to the 
same triangle, they arc similar to each other. 



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192 BOOK III. PLANE GEOMiyi'EY 

Proposition XVI. Tiieohkji 

326. Jf iieo triauffles liavc their }w>nolo<jous sides pro- 
porlional, Ihey ure. similar. 





Given tlie A ABC and A'B'C, in which AB : A'li'^ 
AC: A'C'^BC: B'C. 

To prove the A ABC and A'B'C similar. 

Proof. The given & have their homologous sides pro- 
portional. Hyp. 

Hence it only remains to prove that the A are mutually 
equiangular. Art. 321. 

1. On AB take AF equal to A'B', and on AC take AS 
equalto A'C. Draw Fff. 

Then AB -. AF^AC -. AH. Hyp. 

.-. FH II BC, Art. 320. 

(if a siraiglit line dickies two sides of a A proportionally, il is || the 
third side). 

:. ZAFU = IB, and IAHF= I C. (Why?) 

,•, A AFH and ABC are similar, Art, 323. 

{if tico &. are muttiatly equiangular, they are similar) , 

2. .■- AB : AF=BC : FR. Art. 321, 
Or AB : A'B' = BG : FH. A:^. 8. 
But AB : A'B' = BC : B'C. Hyp. 



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SIMILAK POI-TGONS 

Hence the A AFH and A'B'G' are ecinai. 

.-. A A'B'C is similar to A ABO. 
[for ils equal A AFS is similar to A ABC). 



Proposition XVII. Theorem 
327. If two irianyles have an angle of one equal to an 
angle of the other, and the inciiidhiy sides proportional, the 
trianglea are Kimilar. 




Given the A AHVuud A'B'C, in wiiit^li Z.A^ /.A' and 

AH : A'B' = Aa: A'V. 

To prove tlie A ABC and A'B'C similar- 
Proof. Place the A A'B'C upon t.lie A ABO so that 

ZA' shall' coincide with its equal, the Z.A, and B'O' take 

the position FH. 

Then AB : AP^AG : All. Hyp, 

Hence FII \\ BO. (Why?) 

.-. Z AFII = /.B, D.n<i ZAEF= ZC. (Wiy!) 

.'. A ABC and AFH are similar. Art. asrt. 

Or A ABC and A'B'C are similar. Ax. 8. 

(). E. D. 

miliirA.i'ii'O', 



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Proposition XVIIl 

328. If iivo tnangles have tK.. ^ 

pp.ndiciilar, each to each, the triangles are similar. 




Given tlip A A'B'C (Fig. 2) with it; 



Theorkm 

S'i'les parallel, or per 



and tliM 



■espondir 



A A'H'ff (Pig. 3) with its sidos -L, 
nf the A ABC. 

To prove & ABC and A'B'C similaf. 

Proof. The A A and A' are either equal or supple- 
mentary, {foi-llie»i<lc3fm-miig tJ,emn:-e\\ or X). Arts. 130,13?. 

Similarly, the A B and B', and C and C are either 
eqnnl or supplementary. 

Hence one of the three following statements must be 
true eoneerning the angles of the A: either 

1. The A contain three pairs of supplementary A and 
lA^ lA' = 2vX. A. IB -^ IB' ^2 rt. A, AC + AC' = 
2 rt. A ; or 

2, The A contain two pairs of supplementary A , as 
/.4=ZA', ZB + Zit' = 2rt. A, I C + AC = 2n. A : t,r 

:i. The A contain three pairs of equal A and AA = 
AA', IB=IB'. AC=AC, Art. 139. 

(if two A of a A = two A of another A, the third A are tqnai). 

The first two of these statements are impossible, for the 
sum of the A of two A cannot exceed four rt. A . Art. 134. 

Hence the third statement is true, and the A ABC and 
A'li tj MM liiutuaily cqiiiungular, and therefore Bimihir, 



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Proposition XIX. Theorem 

329. If iiro pohjijmin arc xiMilci-, ihcij may he separated 
into the Sfime number of iriaHglen, similar, each to each , and 
similarly placed. 




Given tlifi similar polygons ABODE and A'li'C'JyE', 
dividml into triangles by tho diagoiiids AC, AD, and A'C, 
A'jy drawn from the corresponding vertices A and A' . 

To prove tliat & ABC, ACD, ADE are similar U) tlie 
^ A'B'C, A'Cry, A'D'E', respectively. 

Proof. 1. /.li ^IB'. Ait.:m. 

Also AJi : A'B'^Tin  B'C. Art. r.'Ji. 

;. AABV and A'B'C are similar, Art. yy.. 

U/Uco ii l,i(fc<iii L of one = <m Z of the i'lhe,- >„;l the ii,<-l,idim; 

2. Again IBGI) ^ IB'C'D', Art.s-i. 

(JiwB(0io(70ns i of similar poliigojin) . 
Also Z BCA = Z i"C".4.'. Art. 321 , 

{homolo!i,ms A of tlie Similar A AlSt: <uid A'H'C). 
Snbtracting lACD = lA'C'D'. An, a. 

But £0 : B'G'=CD : C'Jf, Art. 31!i. 

{homologous sides of siiiiiltir pnJi/goiis are proportioTial). 
And TiC: /i'C = ^lC: ^'G', Art. 321. 

(iMjijr homologous siili^s of the similur & ABCmd A'B'C). 
Hence jIC : A'C ^ CD -. CD'. Ax. i. 

:. the ^ A6'/J and A'G'iy are similar. An. 327. 

3. Ill like manner it oau be shown that the A .iDE 
i^nd A'D'E' are similar. q. z. b, 



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BOOK in. PLA-NK lir.i: 



PROK)tiITIOX XX. THEDliEJl (CONVKKSE or PliU!'. XIX) 

330. // two poli/gons are cotiiposi'd of the name, nmnhvi- 
of triangles, similar, each to each, and similurly placed, the 
polygons are similar. 





Givea the two polygons ABCDE and A'B'OB'W, m 
■wliieh the A ABC, ACJ>, AD-E are similar, respeetivelj", to 
the A A'B'C, A'C'JV, A'lyS', and are smiilarly plaued. 

To prove the polygons ABCJDE and A'B'C'D'E' similar. 

Proof. /.B = ZB', Alt. 321. 

{hfiiKj homologous S of similar &.). 

Also ABCA ^IB'CA'. 

And lACD^lA'C'D'. 

Adding PC(iials. ZBOl) = I B'C'iy. 

In like manner it may be shown that Z Cl>E~ Z (■'!>' ?J\ 
ZBAE= IB'A'E', etc. 

Hence the polygons are mutually equiangular. 

., AB BG ,, . ., ^ . 

^^^'^ .IIP, " "STT^ (homologous sides of mi 



{Why?) 
(Why-, 
(WLy.^ 



Art. 321, 



Again - 
In like n 



BC 
 B'C'~ 



CB 
CI)'' 
(Why?; 



A'B' B'C 
BG_^AC_ CD ^AG 

^ B'C A'C" C'ly A'O 

CD ^BE __^A E 
CD' D'E' A'E'' 
Hence the homologous sides of the given polygons an 
proportional. 

.-. the polygons ABGDE and A'B'C'D'E' are similar. 

Art. 321 
9. X. B. 



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I'ROl'OR'J'lOSAL LINES 

331, Note, It is often eonvsment to write a saries of ( 
ratios, like those uspd ia the above proof, us follows: 

AB^^BC _/'AC'\ _ £D.^('£0^_ DE _AR 
A'B' B'C \A'C'J~ azf yA'D')~ D'E'~ A'E'' 
a ratio which ia used merely to show the equality of two other r 
being inclosed in parenthesis. 



Proposition- XXI. Thkokem 

332. In any triangle-, the bisector of an angle divides 
the opposite side into segments lekich are proportional to the 
other two sides. 



Given the A A H(\ with the line A l> bise^^ti[lg the Z BA C, 
Jitid meeting BC at /', 

To prove ftC ; !>H^A<! -. AH. 

Proof. DriiW the line CF\\AD, and moetiiijc ^1 />' pro- 
dnee,d at F. 

Then 1)€ : DB^AF : AB. Art, 317. 

But Zr^lp'. Art. 124. 

And ls= £p. Art. i:!ii. 

Also £p'— Lp. Hyp. 

.-. Lr= Is. (Why?) 

.-. A .40Fi8 isoseeles, a.m\AC=AP. (Why?) 

Buhstituting .-1 ('for itsec|uiU, A-F, in the above proportion , 

DC : DB^AC: AB. An-, s. 

Q. E. B. 

Ex. If. ill tlie above figure, AB^U.AC- VI. nud /ir r M, find DC 
and Di:. 

[SvG. Let lJC = Jr. /V/f-U — 3, etc.] 



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)S HOOK .III. I'LANi; (;eu.metky 

333. A line divided internally Ua Vuiv divideil into two 
irtK by ii Doint ty.ken betvveen its extremities. 



Tims, tlie line Ali is divided intpriially at I 
into tl)e segments PA and PB. 

To divide a givi'n line interjially in a frU-Pii ratio (as ' 



334. A line divided externally is ii line wliose parts ai 
considered to bo llie segments included between ;t ]ioiiit (j 
the line produced and the extremities of the gi^■cu line. 



Thus, the line AB is divided externally at the point. /" 
into the two segments P'A and P'B. 

To divide a given line externally id a given ralin (us 2 : T), divide it 

ratio (bkT' — 2, (ir .) fiarts) uiid ]iiMdiu;e it lill liic pmdutfd pait ciiuals 
tlie siualler ti.rm of llie ratiu ihiK^ ilie unit line IVmud. 

335. A line divided harmonically is a line divided Imih 

inlprnally aud exUTiuilly iu the same ratio. 



Thus, if the line AB is divided internally, so that PA .- 
PB-l : 2, and externally, so that FA -. /"/>'=! ; 2, then 
PA : FB = PA  PB, and the line is divided harmonieally 

Ex. 1. Divide a givea line barmonicallv in the ratio I ; 3. 
Rs. 2, Divide a given line liarniaiiiciilly in ilio ratio ^^ : j. 



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rKOPORTIOKAL LINES JLH) 

Proposition XXII. Theokem 

386. In any triangle, the hisector of an exterior angle 
diriihs rxiernalhj the opposite side prodiiird into segments 
which arc proportional to the uther two sides. 




Given the A AUG with its sxtei-Lor Z UAO bisected by 
le line AD ineetius tbe opposite side produced at D. 



To prove 




1)B: DC^AB: AC. 








Proof. I 


ru\ 


- the line BF II AD and meeting 


AC at F. 


Tliyii 




DB: OC^AF: AC. 






(Why!) 


But 




Z/' = Z/. 






fWbyr) 


Aud 




Z. = Z/.. 






(Whj!) 


Also 




lp'=Zp. 
.: Zi-= Is. 






(Why?) 
(Why!) 




A 


ADFia isoseeles, and AB 


= AF. 




{Why}) 


Snbstiti 




: AB for its e(inal, AF, 


11 the 


al) 


ve ]>vo- 


ortiou. 




DB -. DC^AB : AC. 






i\\. S. 



337. Cob, The lines bisectttxj the iiUn-ior and ixtcrhr 
anf/lis of a Iriaii'jk at a given vertex, divide- the opiiusite 
siiJt karmoiiicully. 



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Proposition" XXIII. Theorem 
338. The homologous altitudes of two similar ti-ian'jlet 
have the same ratio as any two homologous sides. 




Given the similai- A ABC, A'B'C, und IW, B'J)' 
any two homologous altitudes in these ife. 

B I) B A BO A a 

To prove WD' = WA=Wo'^Tr/ 

Proof. In the rifrht A ABD and A'H'iy, 



Hij hiiiimhigiius A of tbesiniilfir & JlWanil A'B'<''). 

:. A ABI) aiul A'B'I)' are similar, Art. -.i-^A, 

A hace an acuta L of one — an acute I of the other, the & 



■• B'jy B'A'' ''^''^' ■'"'■ 

1 the similai' A ABC and A'l.i'C, 
BA BC A C 

Wa'^Wc'^a^' ^"■^^'■ 

b i> ^ ba ^bc ^a c _ 

b'd' b'a' b'c a'c' ^"^ ^' 

<). E. ». 



(, AC=18, A'C = 12, and tlD = 



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I-ROPORTIONAL LINES 201 

Proposition XXIV. Theorem 

339. If three or more lines pass through the same point 
and intersect two parallel lines, they intercept proportional 
segments on the parallel lines. 





Given the transversals 0^ , OB, OC, 01) intei-secttnE the 
parallel lines Ali and A'D' in the points A, B, C, D and 
A'. B', 6", !>', respectively, 

Ali B a (J n 

To prove a1^'=JF(J'^c1^- 

Proof. A'D' 11 .1 P. R.rp. 

Therefore the hiise A. of the A A'li'O, B'C'O, et,e., are 
equal to corresponding base A of the ^ ABO, BCO, etc. 

All, nti <n- Art. [2.1. 

Hence the A A'B'O, B'C'O, etc, are similar to the A 
ABO, BCO, etc., respectively. Art. 324. 

 iVB' Kii'o) B'V \V'Oj C'l)' ^ '■■ 



Es. If tbe skies of a polyt,'oii 
111* polygon, the side homologous 
the second polygon. 



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MJ, BOOK III. I'LASE GEOJIKTllV 

Prop. SXV. Theokem {<Jonvekse of Prop. XXIV) 

340. // three or more non-parallel straight Unes inter- 
ccpi proportioHul si'/jmrnts oh two pnraUds IJiey pass through 
the name point [thiiC -is, are coiu-iirreiil). 




psirallel lines AC and A'C f 



 that - 



Given the transversals AA', BB\ CC intersecting the' 
AB ^ BC 
' A'B' B'O' 

To prove that the lines AA' , BB' , CC are eoiicurreut. 
Proof. Pt'odnoe the tines AA' and BB' to meet at some 
point 0. 

and let it intersect t 



)nnv the lir 
B point r. 



 A'V Al 



1 H 



IW 



But 
HeDue 



A'B' B'P 

A'B' B'C 

B'1'=B'C'. Art, ; 

.'. point P coincides with point C. 

.: line CC coincides with line CP. Art. 

:. the line CC, it produced, passes through 0, 
(for it coincides with the line CP, iBliich passes through 0). 

:. AA', BB', CC meet iu 0. 



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PEOPOBTIOXAL LlXh;s 203 

Proposition XXVI. Theohem 

341. The perimeters of two similar pohjijotis have ike 
same ratio as any two homologous sides. 





GiveE t Iw two Jiimiliir i">lyt,^uLi.s A IIUDE and A'/i'VITI!', 
with their perimeters denoted by P and P' Slid with AB aad 
.I'/i" itny two homologous aides. 

To prove that F -. r = AB : A'li'. 

Proof, AH  A'H'^BC : li'V^CD : G'I)'^<-t<i. Art. 321. 

Hence An+IW+Cn+,i-tv.. : A'fr+ B'V'+V'I>'+,etc., 
■^Ali : A'B\ Art. 312. 

(iii a wnra of rt/ml rirfio.'^. Ilin s,:,„ <,/ ,<n liir t,:lri-e<lenl>! !« /" Ihrn^im „J 
<\ll Ihc eomequcnls as amj m,t ,u,UT.cilcnl is to its cons<:>pieuf). 

Or P:/" = -t/i ; A' !S' . 

Q. E, D. 

341 (CI. Col;. !i, tiro :<i mile,  'j.ohii/on--^. uiiij two huuwl- 
iKjoax lini-i air hi rack other as iinij other lii'o hoiiioLoijoiis liutx; 
and the (xriiwter^ are to each other asainj two homologous lii,(n. 

Ex. If the porimuUT of a dv.'ii li.'ld is 210 ridf^ ami i. siilr «t ihia 
Held is ti. !L himioloHdus Ad<- of a similar li.Od iii ^ ; 2. tioil Uie i>crmie- 
ter of the iouiucl litlil. 



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Proposition XXVll. Theorem 
342. In a right trianijle, 

I. The altitude to the hypotenuse is a mean proportional 
hetween Die segments of the hypotenuse; 

II. Each leg is u mean proportional between the hypote- 
nuse and the segment of the hypotenuse adjacent to the given 
leg. 




Given thfi Hglil. A ABO and BF ihe altitude upon the 
hypoteniiMe A G. 

To prove !. .IF: HF^HF-. FC. 

I A<! : An = AH: AF 
\ AC : BC^ liC: FC. 

Proof. Th(! Z A is coiriiiinn to the rt. A -4 BF and ,4 BC. 

.: A ABF and ABC are similar. Art. :i*J4. 

Similarly Z <7 is common to the rt. A BFC and ABC, 
and A BFC and ABC are similar. 

.-. &. ABF" and BFC are similar, Art. 3-25. 

{if two & are similar to the sam« A, they are similar to each other). 

Hence, I. In tlie A ABF and BFC, 

AF : BF-= BF : FC, Art. a21. 

{homuhfjin'f suies of simitar & an: pruporUonal), 



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PROPORTIONAL LIXES "JUS 

n. In the simiiar ^ ABC uiid ABF, 

AC : AB = AB : AP. Art. 321. 

Also, ill the similar A ABC and BFC, 

AC:BC=BC: FC. Art. 321, 

Q. E. B. 

343. Cor. The pet-pendicular to the diameter from any 
point in the eircumfercnce of a circle is  
poriional ieiween the segmeiiis of the di- 
ameter; and the chord joining the point to 
an extremity of the diameter is ,a i 
proportional heltveen the diameter and 
the aegmcnt of the diameter adjacent to the chord. 

344. Def. Tbe projection of a point upon a line is the 
foot of the perpeudicular drawn from the point to the line. 




Thus, if AA' he perpendicular to LM, A' ia tiie projec- 
tion of the jHiint A on the line LM. 

345. The projection of a line upon another given line is 
that part of the second line which is included between per- 
pendiculars drawn from the extremities of the first line 
upon the second line. Thus, the projection of AB on LM 
isA'B'; of CD on PQ. is CT). 

Ea. 1. IE the si!|-mentB of tho b.V[ioteimae of a right triangle are 3 
and 12, find the ultitiide on tlie hypotenuse ; also find tha legs of the 
triangle. 

Ex. 2. If, in the figure of Art. ^43, the ciiainetet is 20 and the longer 
chord IK, ti(]ii the hegiiicnt of lliu diu.mets;r udjucent to the chord. 



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Pkopo-^itiox XXVIII. Theorem 

346. In a rigid inawjli-, the square of the hypol'-na^i 
equal to the sum of tlir K(;'(«ivx of the Imjs. 



Given the right trl 



,IC thi' liypot^TiuPfl. 



Proof. Draw the line BF ± AC. Tlieo 

AC : AB^Ali : AF :.AC X AF= AI?. Arts 
Also AC: nC=nC : F<: .-. AC X FC=BC^. 
Adding equals, At' (AF-+ FC) ^Ti? -rluf . 
Or 'aJ^ = AJr + luf. 



347. Cor. 1. 7n a right triang!'', the square of '■ilher 
leg U equal to the sqi^are of the hi/poti-nuae mliii<.s the aquari' 
of the other leg. 

348. Cor. 2. In the square ADCf>. the diagonal di- 
vides the square into two right ti-iangles. 

Henee AC- = AB' + 'bC'=2 AJT. 

.-. AC^^ABV2, '>r~ = '~- 

Hence flte diagonal and the side of a 
square are incommtnsurahie. 



Ex. 1. If the legs of a 
Ei. 2. In tiie figure o 



t. A are 1 J- in. and a in., liiid the liypotenuaB. 
p. 204, show that ~AB- : BC^ = dF : FC. 



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_Vi;MEIiICAL fROlEiri'IKS -07 

Proposition XXIX. Theorem 

849. In any oblique triangle, tlte square of a side oppo- 
xUp an acute angle is equal to the sum of the sguares of the 
other two sides, diminished by twice the product of one of 
those sides hj the projection of the other side upon U. 





Given neiite ^ (7 in A ABC, and DC tlie iii-ojet 
the side BC on the side A C. 

To prove lB-=BC^+AXf — 2 AC X DC. 
Proof. If P fulls oil AC (Fi^. 1), AI)==AC—r)C. 
If Zt falls on ilCpi-odueed (Pig. 2), AD^DC—AC. 
In either case, Air^AC^ + IHf- — 2 ACX DC. ax. i. 
Adding Blf to each of these equals, 

I7r + £Z*'=l6'' + 7*t'' + !B7r— 24CX dc.k^.i. 

But, in thei-t. A ABD, Air + 1^)^ = ~AB" , An, -.m, 

and, in the rt. ABBCIKr+Bir^'BC^. (Wbyf) 

Substituting these values in the above equality. 

AB^ = mf + ACP — 2 AC X BC. ax, 8. 





1, A line 10 i 

ind tlio projeetioi 


Q. E. K. 


Ex. 

line; i' 


[1. loii« ninkes on nnsle of 45" with a seeond 
;i of tli6 first line on the senoiid 


Ex. 


2. Find the 9a 


me, if the angle i» tiO°. 


tion ot 


3, If the aide i 
1 the base. 


jf an equilateral triivngle is a. find it'f projeo- 


Sx. 


4. If, in I'ig. 1 


, iX'=10, AC = 1':, ttud 6C = ^d% lliid.-!«. 



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PliOl^UWlTION XXX. 



350. In any obtuse triangle, tJie square of the side oppo- 
site OH ohtuse angle is equal to the sum of the squares of ike 
other two sides, increased hy twice the product of one of the 
' ' by the projection of the other side «j)oh that side. 




sides 



Given the obtuse lAGB iu the A ABC, and CJ) the 
projection of BG on AC produced. 

To prove AB^= Jo*+"EG^+ 2 AC X CD. 
Proof. AD=AC+ CD. As. 6 

.-. A3'='Ad'+'CB^+2ACXC}). A^. i. 

Adding BI> to each of these equals, 

AI)~-\-inf=^^+'CJJr+'BD^+ 2 ACX CD. Ax. 2. 
But, ill the rt. A ABD, AI?+ lw^= AB'-. (Why f) 

And, ill the rt. A CBD, CI?+ 'bd'= 'BCT. (WLy •<) 

Substituting these values in the above equality, 

'A}?'='A<f+'B<f-\-2 ACY. CD. Ax. 8. 

Q. Z. S. 

361 . Cor. // the square on one side of a tnangle equals 
the sum of the squares ott the other two sides, the angle oppo- 
site the first side is a right angle; for it cannot be aeute 
(Art. 349), or obtuse (Art. 350). 

Et. 1. If, in the above ."igure, BC=10, AC = 2, and IB€A = 120'', 
find AB. 

El. 2. UAB = 2I}, ISC=U, iind AC = 12, find CD. 



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NTiMERIGAL I'EOPEItTlES 209 

Proposition SXXI. Theorem 
352. J/, in any triangle, a median be drawn to one side, 

I, The sum of the squares of the other two sides is egual 
to twice iJie square of half the given side, increased hj twice 
the square of the viedian upon that side ; and 

II. The difference of the squares of the other two sides is 
equal to twice the product of the given side by the projection 
of the median upon that side. 




Given the A ABC, AB>AO, AM tlie median upon BC, 
and MF the projection of AM on BC. 

To prove I. Air+ lc^=2 32?+ 2 'AM^. 

11. Is — Ic^=2 BC X MF. 
Proof. In the A BMA and AMC, BM^SIG. AM^AM, 
&ndAB>AC. Hjp. 

.-. ZAMB k gresiter thun /-AMC. Art. 108. 



,". ZAMB is obtuse {for His greater ilian half a straight Z), 
luohtme £S ABM, A^ = 

In acute A ACM, AG^.^MO^+AM^— 2 . 

Adding, Jb'+Ic'=2 BM''+ 2 AM'', 

{forMC=im). 

Subtracting, Tlf-^AX? = 2BCX MF. 
Ifor BM+MC=BC). 



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PLANE G eOJlETRY 



353. Formula for median of a tiuangle in terms of its 
sides. Ill the Fig., p. 209, deiiothij; AB by c, AC by b, 
BG by a, and AM by )», by Art. 352, b'^ -\- 'j^ ^ 2 m^ 



-K?)% 



'■^(/y^ + . 



354. If iu-o chords in i 
the segments uf one chord i 
inmts of the other chord. 



XXXII. TUEOEKM 

a circle intersect, the product of 
rqiial to the product of the seg- 




Given the O ADBC with Uie chords AB and CD inter- 
eectics at the point F. 

To prove AF X FB= OF X FJ). 



Proof. 


Dm 


y AD and GB. 




Then, 


in the & AFn ami CFB. lA- IC. 
(mk/i hcinfj jNeHSMra? hij i it:-/: IHI). 


Art. 2r,». 


Also 




II)=IB. 


(Wlyt) 


Hcuoe 


the 


& AFD and C FB are similar. 


(Why f) 






.-. AF: CF.FD: FB. 


{Why I) 


And 




AFXFB=CFX FD. 


(Wbjt) 
Q. £. S. 



855. Cor. 1. If through a fixed point within a circle a 
chord be drawn, the product of the segments of the chord is 
constant, in whatever direction the chord be drawn. 



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PROPORTIONAL LINES 211 

356. Def. Four directly proportional quantities are foar 
quantities in proportion iu such a way tliat both, the ante- 
cedents, bolong to, one figure and both the consequents to 
another figure. 

Four reciprocally proportional quantities are four quanti- 
ties in proportion in such a way that the means belong to 
one figure and the extremes to another figure. 

357. Cob. 2. The, segments of two chords intersecting 
in a circle are reciprociUy proporlional (tiie segments be- 
ing considered us parts of the chords, not of the A). 

Proposition XXSIII. Theorem 

358. If, front a given point, a secant and a tangent be 
drawn to a circle, the tangent is the mean proportional he- 
tween the whole secant and its external segment. 



Given AB, a tangent, and AC, a secant, to the cLrele 
BCF, and AF Vad external segment of the secant. 
To prove AG : AB^ AB -. AF. 

Proof. Draw the ehords BC and BF. 
Then, iu the A ABC and ABF, Isi ^ lA. Ident. 

IC= I ABF. Arts. 258, 264. 

.-. the ^ ABC and ABF are .'^iiiiihu-. (Why n 

.■. AC : AB = AB : AF. (Why !) 



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J-U HOOK III. PLANK GF.OMKTKY 

SbQ. Cor. 1. If, from a given point, a tangent and a 
secant he drawn to a circle, the product of the wJiole secant 
and its external segment is equal to the square of the tangent. 

860. Cor. 2. If, from a gi fen point without a circle, a 
secant he drawn, the product of the secant mul its external 
segment is constant, in witatever direction the secant be drawn. 

Fop the product of each secant and its external seg- 
ment eqnnls the square of the tang'ent, which is constant. 

361. Cor. 3. If two secants be drairii from an exter- 
nal point to a circle, the whole secants and their external 
segments are reciprocally proportional. 



pROPOPiTiox XXXIV. Theorem 

362. The square of the bisector of an angle of a triangle 
is equal to the, product of the sides forming the angle, dimin- 
ished by the product of the segments of the third side formed 
h'j the bisector. 




Given the A ABC, CF {or t) the bisector of ZACB, 
aad m and n the segments of AB formed by CF. 
To prove t^ — ab — nin. 

Circumscribe a © about the A ABC, 



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KUMEKICAL PISOT'EETIEa 



Proof. Prbduee OF to meet the circumference 


at H",.aud 


di-aw the chord BH. 




Then, iu the A ACF mdi CHB, AA = III. 


(Why ?) 


And IACF= IHCn. 


(Why !) 


Hence the A ACF and CUB are similar. 


(Wby?) 


.-. 6: £ = ic + (:a. 


(Why ?) 


.-. ( {xArt)=al. 


(Why!) 


Or (a- + /2 = ai. 




Subtracting ix, f = (ih — tx. 


Ax. 3. 


But /;c = WH. 


Art. 3r)4. 


Substituting liin for te, 





363. Formula for bisector of an angle of a triangle. 
m:n = h :«(Art.332).-.m + K:m = i + (( ; (.{Art. 309), 
or c ; m — o■■^- h -. h (Ax. 8^ 



he 



In like manner K^^ — —r- Substituting for i 

(I + b 
nhr^ _oh(f< +h + r){a +h- 



(^=--ab- 






Ei. 1. Oil the figure, p. 210, lot ^F=10, rR = 4, and FC^S. 
'mil FI>. 

Ex. 2. On thfi flfc-urH, p. 211, let AC = IG ui.d .JC = ia, Find AF. 

Ex. 3. On the figuvo, p. 213, let 4C=1G, CB = 12, and 4C = 14. 



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Zli JiOOK in. PLANE GEOMETHY 

Pkopositiox XXXV. Tuboreji 

364. Jn any triauyle, the product of any two sides is 
equal to the product of the diameter of the circumscribed cir- 
cle by the altiiitde upon the third side. 




Given the A ABC, ABCJ) a circumscribed O, BDthe 
diameter of this O, and Bi^'the altitude upon AC. 

To prove AR X BC=BD X BF. 

Proof. Draw the chord DC. 

Then, in the A ABF and DBG, I A = ID. CWhyS) 

IDCB is art. Z. (Whyf) 

.-. & ATJJi'and 7>/J6'are simihir. (Why!) 

.-. AB : BB^BF: BG. (Wl.yf) 

.-. A B X BC=BD X BF. (Why?) 
q. E. D. 

365. Cor. The diameter of a circle circmnscribed about 

a triangle eqiiah the product of iivo sides of the triangle 
divided by the altitude upon the third side. 



Ex. In the above figure, if AB = 8, CC = U, and ilF = 4^, fled the 
rtidiua o£ tbe eireumBcribed (.'iicle. 



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construction problems 215 

cohstroction problems 
Proposition XSXVI, Problem 
6. To construct a fcurth proportional to three given 



Given the lines m, n, p. 

To construct a fourth proportional to «s, n, iuid p. 

Construction. Take auy two liuea, .4/* and AQ, making 
any convenient /.A. 

On J-Ptake J.B = »k, and BC^n. 

On AQ take AI)= p. 

Draw BB * 

Through the point C draw CB\\J)n, and meeting A^ 
at E. Art. 2T9. 

Then I>Ii is the fourth proportional required. 

Proof. An : BC=AI> : DR. Avt. :siT. 

Or m : n^p : DU. Ax, S. 



Proposition XXXVII. Problem 
367. To consiriiet a third proportional to two given lines. 
Let the pupi! supply the construetion and pr'oof . 
[SUQ. Use the method of Art. :i6C, ninking p = n.] 
Ex. 1. Construct a fourth proportional to three lioea, j, 1 and 
I In. long. 
Ex, 2. CocBtnict a third proportional to two lines, 2 and 14 in. long. 



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BOOK m. PLANE GEOJIETliif 

Proposition SXXVIII. Problem 
To construct a mean proportional between iivo given 




Given the lines m and n. 

To construct a mean proportional between m anil n. 

Construction. On the line AP take AB= m, and BC= 



At B erect a ± to AG meeting the semi-eironinferenee 
at R. Art. 274. 

Then BE is the mean proportional reqnired. 

Proof. AB : BE=BB : BG, Art. 343. 

{Ihc J- to the diameter from atiij points in the eirciniferance of a circle is 
a mean proportioiial belweeii the segmenls of the diamvler). 

Substitnting for AB and BG their values »« and n, 

m : BB^BK -. n. Ax. 8. 

Q. E. F. 



Ex. 1. Construct the third propovtloaal to two I'mes, 1 and IJ in. 
long. 

Ex. 2. Construct a mean proportional between two lines, 1 and 2 
in. long. 

Bi, 3. Taking any line as 1, oonstiniot j/S. 



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CONSTRUCTION fROELEMS 2l7 

Peoposition XXXIX. Peorlem 

369. To divide a givm straight line into parts propor- 
tional to a number of given liiies. 



"^' 



\ 



Given the straight lines AB, m, n aiiJ /!. 

To divide AB into parts proportional to »t, n and j). 

Construction. Draw the line AP, making any convenient 
angle with AB. 

On APmark off AC=»H, CD=n,mdI)F=p. Draw-BP. 

Throngh the points C and D draw lines || BF, and meet- 
ing AB at E and 8. Art. 279, 

Tliou AR, US and SB are the segments required. 

_ , AB ES SB 

^"°^- AC^cB^W ^''■'''- 

{if two lines are cut by a wimbey ofparallets, the corresponding segments 
are iiroportional) . 
For A C, CD and BF substitute their equals »», n and p. 

„, AB B8 SB 

Then  — = — = —-. ax. h, 

m 11 2) 

Q. E, If. 

370. Def. a straight line divided in extreme and mean 

ratio is a straight line divided into two segments such that 

one of the segments is a moan proportional between the whole 

line and tiie other segment. 



Ea, Divide a given line lato parts prupc 



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18 BOOK III. PLANE GEOMETRY 

Proposition XL. Problem 
371. To divide a given straiglii line in extreme and, 




Q 'I 

Given the line AB. 

To divide AB in extreme and mean ratio. 

Construction. At one end of the given line, as B, con 
struct a X OB eqnul to \AB. Arta. 274, 275, 

From as a center and with OB as a radius, describe 
Sl circumference. 

Draw AO meeting the circumference at C, and produce 
it to meet the circumference again at F. 

On A B mark off AP eqnal to A€\ on BA produced 
take A<J = AF. 

Then AB is divided in extreme and mean ratio inter- 
nally at P, and externally at Q. 

Proof. 1. AF: AB=-AB : AC. ah. n5fi. 

.-. AF~An:AB=AB — AG:AG. (Win-?) 

But CF=20B = AB, and AG=AP. 

Hence AF~AB=AP, and AB — AC^PB. 
.: AP:AB = P£':AP,orAB:AP=AP: PB. 

As. 8, Art. 308, 

2. AF: AB = AB : AC. (Why?) 

.-. AF+AB:AF=AB + AC:AB. (Why!) 

But AF+AB=QB. anH AB + AC^AF^QA. 

:. QB: QA = QA : AB. Ax. B. 

Q. £. F» 



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consteuction problems 219 

Proposition XLI. Problem 

S72. Upon a given straight line, to construct a polygon 
similar to a given polygon, and similarly placed. 




Given the polygon ABODE and tlie line A'B'. 

To construct on A'B' a polygon similar to ABODE and 
similarly placud. 

Construction. In the given polygon, draw the diagonals 
AG and AD, dividing the polygon into triangles. 

At /i'^on the line A'/f, eonstruet ZA'B'O' equal to /.B; 
and at A' construct Zll'A'O" eqm\\ to ZBAO. Art. 2TS. 

Pi-oauce the lines B'C and A'C to raeiit at C. 

In like manner, on A'O' construct A A'C'D', equiangular 
with A AOD and similarly placed; au.l on A'D' construct 
the A A'D'E', equiangular with A ADE uud similarly 
placed. 

Then A'B'C'D'E' is the polygon required. 

Proof. The ^ A'B'C, A'C'D', etc., are similar to the 
& ABC, ACD, etc, respectively. Art. n24. 

Hence the polygons A'B'C l>' l-l' and ABCBE are 
similar. Art. :iJO, 



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PLANE GEOMETKV 



EXERCISES. CROUP SO 

SDllLAR TiilANGLES 

Let the pupil make a list of all the conditions that ? 
tivo triangles similar (see Arts. 323, 324, 325, etc). 

Ex. 1. Given AD ± BC, and BF 1. AC; prove 
& ADC and BFG similar. 

Ex. 2. In the samG figure, prove the 
and BFC similar. What other triangle 
figare i3 similar to A BFC f 

Ex. 3. Two isosceles triangles are si 



theii 



 uqual 




Ex. 4. Two iBOBGeles triangles are Bimi 
equals a base angle of the other. 

Es. 5. Given arc -iC=aro J!C; prove A 
AFC and AFC similar. 

Ex. 6, Prove that the diagonals and bases 
of a trapezoid together form a pair of similar 
triangles. 

Ex. 7. AB is the diameter of a circle, BD is 
a tangent, and AD intersects the clreumfetenco 
angles ABE and ADB similar. 

Ex. 8. liC is a chord in a circle, AQ is the di.imeter perpendicular 
to BC and meeting it at 2f; AF is any chord intersecting BC in M. 
Pi'ove the A JJfif and APQ similar. 

Ex, 9. The triangle ABC is inscritied in a circle; the bisector of 
the angle A meets BC in 1) and the circumference in F. Prove the 
triangles BAV and AFC similar. 

Ex. 10. A pair of homologous medians divide two similar triangles 
into triangles which are similar each to each. 

Ex. 11. Two rectangles are similar if two adjacent sides of one 
are proportional to the homologous sides of the otlier. 

Ex. 12. Two circles intersect in the points A and B. AC and AD 
are each a tangent in one circle and a chord in the other. Prove the 
iiJ^C and JBi) similar. 

[8ufl. Prove IBAD= ^ACB, tti!.] 



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EXEBCISES. PROPORTIONAL LINES 221 

373. Proof that lines are proportional. In order to 
prove that certain lines are proportional, or have propor- 
tional relations, it is usually best to show that the given 
lines are homologous sides of similar triangles. 

Sometimes, however, other methods of proof are used 
(as the theorems of Arts. 354 and 358) ; but these, if in- 
vestigated, are usually found to be the method of similar 
triangles in disguise. 

EXERCISES. CROUP 31 

PROPORTIONAL LINES 

Ex. 1. Ontho iignpe of Ex, 1, p. 220, prove JI'X /JC=B-FX--ir', 
and TSC'X Or> = BOXFC. 

Ex. 2. On the figure of F.k. 5, p. 320, prote CP: CA=CA : CF. 
(Hence as P moves tho product of what two lines is constant i ) 

Ex. 3. The diagonals of a trapezoid divide each other into pro- 
portional eegments. 

Ex. 4. In the isosceles triangle ABC, AJi = AC,,oa tho side ^fl, 
the point P is taken so that PC equals the base. Prove AB X PB = m?. 

Ex. 5. In a triangle the median to the base bisects all lines 
parullel to the base and terminated by tlie sides. 

Ex. 6. If PQ is nnj- line throngli F. tl.e midpoint of the line AB, 
and AI' mid BQ are perpendicular to PQ, show thai the ratio PF: FQ 

Ex, 7. The triangle ^CC is inscribed in a circle. J^is the midpoint 
of the arc JC and Bi*" intersects the line 46' in ffi Prove AB : liO = 
AE: EG. 

[SUG. I'se Art. 332,] 

Ex. 8. If two circles intersect, the common ehoid, if produced, 
bisects the common tangent. 

[SUG. Use Art. 358.] 

Ex. 9. If two circles intersect, tangents drawn to the two circles 
from any point iu the common ishord produced ure equal. 



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22'2 BOOK III. PLANE GEOMETEY 

Ex. 10. Givi'Ti AJ>. I'T and EG I ; 
prove PQ = UT. 

[SuG. Show that ^= If', by show- 
ing them equal to a uommon vatio.] 

Ex. II. Lilies aveiltawn from a point within the triangle, to the 
vertices of llis triangle AltC. From B' any point in OB, £'A' is 
drawn parallel to J)'J and meeting OA in A', aud B'C' is drawn parallel 
to BC and meeting OC in C Prove A'li' : AB=B'C' : BC, and the 
ti'iangles ABC and A'D'ff similar. 

Ex. 12. GiyenJ7(c;Dn£I7,and ^ ^ 

F any point in BC produced | prove 
AE' = IIQXI1P- 

[St'G. Compare the similar A 
ABR and BQD ^ also the similar & 
AMD and EBP.'] " " 

eXEFtCISES. GROUP 85 

NUMF.RICAL PROPERTIES OP LINES 
Ex. 1. If AD is the altitude of the triangle ABC, Tl^—lc?^ 

luP'—m:"- . 

Ex. 2. If the liingoiials of a quadrilateral are perpendicular to 
each other, the sura of the squ.ares of one pair of opposite sides 
equals the sum of the squares of the other pair of aides. 

Ex. 3. The square of the altitvido of an equilateral triangle is 
three-fourths the square of one side. 

Ex. 4. If AB is the hj-potenusc of a right triangle, and the leg BG 
is hisected at K, ZZJ" - AM? = SOff-. 

Ex, 5, PlJ isaline parallel totbehypotenuse^J of arighttriangle 
^JJC, and meeting JCiuPand^Cine, Prove Z^^ + 5P = Zb^+P^2. 

Ex. 6. Ill the right triangle AhC, BE and Cf bisect the legs AG 
naAdB in the points £ and F. Pvove iR^ + ici^=5B&. 



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PJ^- 



-PC^ 



EXERCISES. AUXILIARY LINES 

EXERCISES. CROUP SS 

ArXILIARY LIKES 

iiven ^Br/i ft veutHnglo; prove 
= 'PI'?-\-'PD-. 



Ex. 2. The 

eles divides tlie line of centers into segments 
which have the same ratio as the diameters 
ol the oircleE. 

[Suo. Draw radii to the points of coi 



Ex.3. 

n altitiidi 



■IB and -■IC are the liigs of a 
Prove 2JCX-F'C = i(6'2. 



Ex. 4. Given the chords AB and CD perpen- 
dicular to each other and intersecting at O; prove 
(W^ + OB^ -r OC^ + 0O^= (diameter)^, 

[SuG. Draw tlie diameter BE and the chords 
AC, BD, DE. Prove AC=ED. etc] 



Ex, 5, ABC is an inacrihed i 
of which AB and AC are the legsi 
Ftov6A^=ADXAE. 










AD is a chord r 



3elos triangle ABC, and D is 






n the t 



6 produced, then CD^=CB- + ADX BI). 



Ex. 7. Two circles touch at the point T. I'TI" and QTO" are lit 
drawn meeting the circumferences in P, Q and P", (/ resiiective 
Prove the triangles PTQ and F'TQ' similar, 

[Sua. Draw the common tangent at T.] 

Ex. 8. Two circles touch at the point T, throngh 3" three lines i 
drawn meeting the circumferences in i', Q, E and P', (/, E', reapi 
tively. Prove the triangles PQR and PQ'Ji' similar. 

Ex. 9, If A is the midpoint of CD. an arc of a circle, and AP 
any chord intersecting the chord CV In Q, prove that APKAQ h 



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2^i BOOK ill. PLANE GEOMET 

Ex. 10. In nn iiiseribed qnadriluterdl, the 
product of the dingonala ia equal to the sum o£ 
the products of the opposite sides. 

[8UG. Drfiw BFfO that ICBF= lABD and 
use similar trian{|;lts.] 

Ex. 11. The sum of the siiiian's of t]iy 
aides of any quadrilateral is equal to the sum 
of the squares of the diagonals, plus four times 
the square of the liue joining the loidpoiuta of 
the diagonals, 

EXERCISES. CROUP 34 

INDIRECT DEJIOXSTRATIONS 

Ex. 1. If the sum of the squares on two sides of a triangle is 
greater than the square on the third side, the angle included by the 
two given sides is an aeute angle. 

Ex. 2. If D ia a point in the side AC of the triangle AJ!C. ajid 
AD : DC = AB : BC, then Dli biseeta angle ABC. 

Ex. 3. A given stmight line can be divided in a gtveu rritio at but 
one point. 

Ex. 4. If the sides of two triangles are par- 
allel, each to eauh, and a straight line be paasiid 
through each pair of homologous vertices, these 
lines, if produced, will meet 

Ex. 5. If each of three c: 
other two, the three common chords intersect in 



EXERCISES. CROUP 35 

THEOREMS PROVED BY VARIOUS METHODS 
Ex. 1, In the figure on p, 204, show that ABy, BF=BCX AF. 
Ex. 2. In the same figure, if J'C=3vli'', sliowthatZS^ : fl(?- = ! : 3. 
Ex. 3. AB is the diameter of a circle and PB is a tangent. If AP 
meets the ciccumfeteneo in the point §, prove that APX. AQ = AB\ 




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MIfiCELI.ANEOTIS EXERCISES. THEOREMS 



Ex. 4i. If the line hisocting the parallel sides of a tra,pezoid b 
produced, it meets the legs produced in a common point. 
[SUG. See Art, 340.] 

Ex. B. In similar triangles, homologous uiediaua have the aam 
ratio as iiomologoua sides. 

Ex. 6. A diameter AB is produced to the point C : CI' is perper 
dicular to JC; PB produced meets the eiruumterence at Q. Prov 
the triangles A QB ai.d POP. similar. 

Ex. 7. If PA and PB are chords in a eirele, and CD is a line pai 
aliel to the tangent at J* and meeting PA and 
PJl at C and I), the triangles PAIS and PCD are 
similar. 

Ex. S. Given AB a diameter and AD and BO 
tangents, JCand DB intersecting ut any point F 
on the circumference; prove AB a mean propor- 
tional between the aides AD and BC. 

Ex. B. In any isosoelea triangle, the square 
of one of the legs equals the square on a line 
drawn from the vertex to any point of the base 
plus the product of the segments of the base. 

Ex. 10. A line drawn through the intersection of the diagonals of 
a trapezoid parallel to the bases and terminated by tho legs is bisected 
by the diagonals. 

[SuG. SeeEs. 10, p. 222.] 




Ex. 11, If a chord is bisected by another ehorl eath segmi 
the first chord is a mean propoitiouil letweei the begui'nts of th 
second chord. 



of 



Ex. 13. If two circle'* aie tangent exteinallj and a linn is di: 
through the point of lunta t terminated bv the cireumferoncea 
ohordu intercepted in the two circles are to ^ath other ai the rain 

Ex. 14. Thz-ee times the sum of the squares of the sides oE a 
Bugle equals (our times the sum of the squares of the medians. 



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2'^(> 



BOOK ni. PrANK 



El. 15. Find the loeua of the jHklpoiuts of 
lines in a trinngle parallel to the biis" and ter- 
minated by the sides. 

Ex. 16. Given AB the iliam^ai-r. AP. PQIi, 
BR, tangents; pmve P^ X Qli a<:ousUnt(= ra- 
dius squared j, 

Ex. 17. is the center of h circle and A is 
any point within the circle ; OA is produced to J), 
so that OA'KOB equals the radius squared. 
If P is any point in the oivcumferonee, the 
angles 0J>^ and OBI* are equal, 

[SUG. Use Art. 327.] 

Ex. 18. Given AF=FI1, and ri{ \\ Jl! : 
prove BP : FP=HK : FK. 

[SuQ. EP : FP=UII: FH, utc] 

Ex. 19. If from any point P within the triangle AJIC the perpen- 
diculars /'§, ^.B, PTare drawn to the sides Jii,JC, BC, re^ p. actively, 

EXERCISES. QROUP 3a 




Ex. 1. Construct J- = — ; ulso .c= — . 

Ex, 2. Construct i- = \/ <i'— iF~, i. a., l/'lTT'f Th"-^. 

Ex. 3. Construct i = V''iai, i. B., "|/(3 u) /i. 

Ex. 4. Construct ■x = \/d' — t>r', i.e., 1^ «" - {l/fiTo'. 

Ex. B. Given a line denoted by 1, eoustiuct i/3; also iv/5. 

Ex. 6. Divide a line inta three parts proportional to 2, \, %, 

Ex. 7. Divide a line harmouieaily in the ratio 3 : 5, 



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KXEKOISES. PROBLEMS !2'i7 

ti'iangie into segments propoi'tioual la 

Ex. 9, Divide aline into segmenta in the ratio 1 ; -j/S. 

Ek, 10. Given a point P in the aide JB o£ a triangle ARC, draw 
a line from P to AC produeeii so that the line drawu niiiy be bisected 
byBC. 

[SUQ. Suppose the required line, Pqp., drawn meeting liC in Q 
andJCinB. From P draw Pi H JC. Compare the APL^ and yjiC] 

Ex. II. Through a given point F in the are 
subtended by the chord AB draw a ohord which 
shall be bisected by AB. 

[Suo. Suppose the required chord drawn, viz., 
PQR. Jointlieeeoter O with P and y. What kind 
of an angle is OQP, etc,?] 

Ex. 12. Ill an obtuse triangle draw a line from the vertex of ths 
obtuse angle to the opposite Bide which shall be a mean pioporticmal 
between the segments of the opposite side. 

[Sua. Cireumscribe a eirole about the triangle and reduce the 
problem to the preceding Ex.] 

Ex. 13. Find a point F in the arc subtended by the chord AB 
such that chord PA : chord PB = 2 : 3. 

[SuG. Suppose the required construction made, and also the ehnrd 
.^B divided in the ratio 2 ;3 at the point Q. How do the angles Ji'g 
and QPB compared] 

Ex. 14. Given the perimeter, conatruet a triangle similar lo a 
given triangle. 

Ex. 15. Given tliH altitude of a triangle, construct a triangle 

Ex. 16. In a givun circle inscribe a triangle similar to a given 

Ex. 17. About a given circle circumscribe a triangle similar to a 
given triangle. 

Ex. J8. By drawing a line parallel to one of the sides of a given 
rectiiugle, dividt thu rcclanaio imu Iwo ainjilar rectangles. 



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228 



BOOK Hi. PLANE (lEOMETUY 



Ex. 19. InKoribfl a square in a given 
triangle. 

[Si'O. If ABC is the given tnaagle, sup- 
pose hOFE the required inaeribtd square, 
.ioin HE and produce it to meet AH \\ IIC. 
VroveJH=JK, etc.] 

374. The metJiod of similars in solving geometrical 
problems is best shown by the aid of 

Ex. In the side JSC of a triangle JHt 
tlie perpendiculars from it to tho othBr 
sides shall be in the ratio 3 : 1. 

CoNsTRUCTiOH. At any point /' in J'' 
erect ft X PQ of any convenient length, 
lu a direction ± All draw Jfy= i Ql'. 
ioia.EP. V-coiaSiTaw tiT\\IlQ. Produce 
AT to D. Then D is the required point. 

Let the pupil supply tiio proof. 




EXERCISES. QROUP 3? 

PROBLEMS SOLVED BY METHOD OF SIJIILARS 

Ex. 1. In one side of a triangle find n point such tliat the perpen- 
diculars from it to the other two sides siial! be in the ratio «i : n. 

Ex. 2. Find a point the perpendiculars from which to tlio three 
sides of a given triangle shall be in a given ri 

[SUG. Use Ex. 1 twice.] 

Ex. 3. Construct a oircle 
which shall touch two givon 



lin 






thr. 




given point. 

[SUG. Let OA and OC bo 
the given lines and P the 
given point. Draw any OR toueliing the two liups {OB at r} and 
intersecting OPprodueed at X. Dtaw the chord A'F, etc.] 

Es. 4 Inscribe a. square in a given semi-circle. 

[Sua. Circumscribe a semi -circle about any given square, by taking 
the midpoint of the base of the square as a center, and the line from 
this midpoint to a uou-adjaeent vertex as a radius, etc] 

Ex. 5. Solve Ex. Ill, p 228, by the method of aimllara. 



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KXEKCISl'lS . PROBLEMS 



229 



375. Algebraic analysis of problems. The cmuUtions of 
a problem may ofitn be stated as an algebraic equation; by 
sohiiig the equation, the length of a desired line in terms of 
knotcH lines may then be obtained, and the problem solved 
by construeting the algebraic e^^ression thiis obinined. 

Ex. Find B point P in 
the line AB mah tLat 'Jf-= 
IBi-'. " 

Analysis amd Construction. Denotu AB by ; 
and PB\>ja — x. Tlien K' = 3{a — a;)'. 



Construct a v'3, whanee construct '— — ; lay ofE the line ob- 

tained, 88 AP, on AB; this gtv&s the point P of internal division. 
Similarly, the eonatriiction of   -  ^ — gives F' , the point of external 



3ISES. CROUP 3S 



. SOI.VKD HY ALGI'IBBAIC ANALYSIS 



E. tbatJ/'-=2BP-. 



Ei. 1 . Find a point f in a given iii 
En. 2. Construat a riglit triangle, given one i 
a and the projection, 6, of the other k-g on i 

[SuG. Denote the projection of a on the hy| 
tennsebyai. Then o= = a: U' + 6), etc.] 

Ex. 3, Inscribe a square in a given aemicircle. 

Ex. 4p. From a given line out oft a part which shall he a mean 
proportional between the rumaiuder of the line and another given line. 

Ex. 5. Given AC and CB arcs of ilO°, and « a 
given line. Draw the ehoiii CQ intersecting AH 
in J'aothat7'y=(i. 

[Sto. .c=-j' = y'. <r.i: = (r-i'i,) (/--J), elc] / 

Ex. 6, Given the greater segment of a line di- 
y\C -i extreme and mean ratio, construct the line. 




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Zdii BOOK III. PLANE GEOMETTiY 

EXERCISES. CROUP 3Q 

PROBLEJIS SOLVED BY VARIOIIS .MF/I'HODS 

Ex. 1. Conatruet two lines, given tljeir Hum [a line AB) and their 
ratio {m ; «). 

Ex. 2. Construct two lines, given tiioiv diffovencK iind tlieir ratio. 

Ex. 8, Divide a trapeaoitt into two sitoilur tviifitKoids by drawing 
a liue parallel to the bases. 

[Suo. Coneeive the fisiire drawn, and compare the ratio of the 
bases in the two trapezoids formed.] 

Ex. 4. Construct a mean proportional between two given lines by 



Ex. 6. Prom a given point draw a seeant to a eircle so (hat the 
extprnal segment shall equal half the secant. 

[SuG. Draw a tangent to the O and use the algebraic method.] 

Ex. 7. From a given ©sternal point P, draw a secant meeting a 
circle in A and B bo that FA : AB= m : n. 

[SuG, Draw a tangent to the circle from the point P and denote 
its length by *. Denote PA by inx and ^S by lij-. Then in{m-\-ii)x' 

Ex. 8. Through a given point P draw a straight line so that the 
parts of it, incliKied between that point and perpendiculars dvawn to 
the line from two other given points, shall be in a given ratio. 

[SuG. Join the last two points, and divide tlie line between ttiem 
in the given ratio.] 

Ex. 9. Conatruet a straight line so that the perpendioulacB on it 
from three given points shall be in a given ratio. 

[Sue. Let F, Q, R, be the given points and m : n : p the given 
ratio. Divide FQ in the ratio m : n and Qlt in the ratio n : p, etc] 

Ex. 10. Upon a given line as hypotenuse construct a right tri- 
angle one leg of which shall be a mean proportional betwf - tlie 
other leg and the hypotenuse. 



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Book IV 

AREAS OF POLYGONS 

376. A uait of surface is a square whose side is a unit 
of length, as a square inch, a square yard, or a square 
etutirneter. 

377. TIip area of a surface is the number of units of 
surface which the given surface contains. 

It is important for tbe student to grasp firmly the fact that area 
menus not mere vague largeness of surface, but that it is a nainber. 
Beinf^ a number, it can be resolved into faKtors, it may be determined 
as a product of simpler numbers, and handled with ease and precision 

378, Equivalent plane figures are plane figures having 
pqunl areas. 

Thus two triangles may have equal areas (he eqiiivaletit) 
and ret not be of the same shape, that is, not be equal 
(congruent). 

379, Abbreviations. Instead of "area of a rectangle," 
for example, it Is often convenient to say simply "rectan- 
gle." So instead of "the number of linear units in the 
base," we use simply "the base." In like manner, for 
"product of the number of linear utiits in the base by the 
number of linear units in the altitude," a common ab- 
breviation is "product of the base by the altitude." 

(231) 



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-•J-i BOOK lY. PI,ANE GEOMETRY 

COMPARISON OF RECTANGLES 

rROPOSiTiON I. Theorem 

380. If Itco rectimijles have the same altitude, llinj < 
to each other us their lases. 
















Given the rectangles EFGJT and J7>CI>, having their 
altitiideri l-'Fand Ali equal. 

To prove EFGB. -. ABCJ)^ hW -. A 7). 



Case I. 'SVIien the hanes are (■ommensumNe. 

Proof. Take some eomraon measure nf fT/f and .1/), as 
AK, and let it be contained in EH n times and in Al) vi 
times. 

Hence EU -. AD = n im. (WLy?) 

Through the points of division of the hases of the two 
rectangles draw lines perpendicular to the bases. 

These lines will divide EG into h, and AC into m small 
rectangles, all equal. ^"- "'^' 

Hence EFGR : ABCD^m m. {Wh;?) 

.-. EFGS ■.ABCD = EU: AE. (Why!) 



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COMPARISON OF RECTANGLES 233 

Case II. When ike bases are incommensurable. 



A D E Qil 

Proof. Divide the base AD into Jiny numher of ftiiinL 
parts, and apply one of these parts to EU. 

It will be contained in EH a certain nnmber of times 
■with a remainder QU, less than the unit of measure. 

Draw gP ± EH, meeting FG at P. 

Then EQ and AI> are commensurable. Oonstr. 



. EFPQ _EQ 
 ABCD A3 

If now the unit of measure be indefluiteSy diminis 
the line QH, which is less than the unit of measure, 
be indefinitely diminished. 

.-. A-(>^£'/fu--alimit; EFPQ^^EFGHi>.i.i^\im\L An 
EFPQ T^_,^„,„^ ^ „„,.ifl>,v wlch ^^^^ . 



Case I. 




381. Cor. // two rechinijUs have e(piul bases, iheij i 
to each oiher ai> their ullilutks. 



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264: ROOK IV. PLANK OEO^tF,TKy 

PliOl'OSITION II. TilliUUKM 

3S2. Tl)it ureas of any two reciani/hs are lo aii:h other 
as till- proOiirta of their bases iij their altitudes. 



Given the rectangles E and R', iiaving the bases b auiI 
6', and the altitudes « and «', respectively, 

_ R h X a 

^'^''"'' R=VX7^' 

Proof. Constrnet a rectangle, N, liaviug iUs ba^^e equal 
to that of R, and its altitude equal to thai, of B'. 



Also 



R'' 



Taking the produet of the corresponding members of 
the two equalities, 

Q. E. D. 



Ex. J. Fiud the ratio o 
&re 12 X 8 iu. to that of o: 



Ex. 2. How many brieks, eacL 8X5 
avementeOX 9"-* 



of a rectangle whose di 
will it take to 



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AKEAK OF POLYGONS JoO 

AREAS OF POLYGONS 

Proposition III. Theorem 

3S3. The area of a rectangle is equal to the product of 
Us base by its aUiluile, 



,m 



Given the reotanglefi, with a base containing b, and an 
altitude containing h nnits of linear meaanre. 

To prove area of B=b X h. 

Proof. Let fTbe a square each side of whicli contains ( 
unit of linear measure. 

Then t-'^is the unit of surface. ArL. 'sm. 



1X1 



= bXh. 



— is the ar 


aof B, 


{h<, deniHllmi 


»/(i™n) 


amiof Jf = 


= bXk. 



(See Art. 1.) 

Ex. 1. Find, 
audS ft. wide. 

Ex, 2. The ar 
Fiud the biwe. 



Ase of this theorem, the probiem of liuding tbe 
reduced to tlie simpler problem of meaaurinj; 
ons of the reclaiigle and taking their product. 



i square feet, the 
X of a reeUnglB is i 



ingle 8 yds. long 
,s altitude is 5 ft. 



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PROTOblTION IV. TliEOHE.M 

385. The area of a parallelofjmm if ''<juid to thi; product 
of its base by ils aUitude. 




Given fcho OJ ABCD with the baso AT) (douote.! by h) 
and the altitude DF (denoted by It). 
To prove area of ABCB = h X h. 

Proof. From ^i draw AK || DF. and meeting CB pro- 
duced, at K. 

Tlien AK ± GK. Art. 133. 

.■. AZFD is a rectangle witli base band altitude /(.(Why!) 
In the rt. A AKB and 1)FG, 
AB^BG. 
AK=BF. 
:. A AKB^A t)FG. 
To each of these equals add the figure AliFD; 
Then rectangle A7i'i^O oro ABVD. 
But area of the rectangle AEFI)=b X h. 
.-. nrcaa ABCD= h x /*■ 

Q.E.D. 

386. Cor. 1. Parallelograms which have equal bases 
and equal altitudes are eqmvalent. 

387. Cob. 2. Parallelograms which. have equal bases 
are to each other as their altitttdes; 

Parallelograms ichich have equal altitudes are to each 
other as their bases. 

388. Cor. 3. Any two parallelograms are to each other 
as the products of their bases a7id alliiadtin. 



(WLy!) 


(Why 7) 


(WhyT) 


Ax. 2. 


Art. 383. 


(Why ?) 



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AEEAS OF POLYGONS 3d7 

Proposition V. Theorem 

889. The area of a triangle is equal to one-half the 
product of its base by its altitude. 




Given the A ARC with tlio base AC {denoted by i), and 
the altitude Fli (aenoted by /(). 

To prove area of A ABC^i h X Ji. 
Proof. Draw BD \\ AC, and CD ]] AB, forming the /Z7 
ABDC. 

Then BO is a diagonal of HJ ABJ)C. 

:. A ABC^hCJ ABDC. Art. ir.fi. 

But areaC7A£DC=6 X ft. (Why?) 

.-. area A ABC=i 6 X ft. Ax. 5. 

Q. £. D. 

890. Cob. 1. Triangles which have equal bases and 
equal altitudes {or which have equal bases and their vertices 
in a line parallel to the base) are equivalent. 

391. Cor. 2. Triangles n-htch have equal bases are to . 
fach other as their altitudes; 

Triangles which have equal altitudes are to each other as 
their bases. 

392. Cor. 3. Any two triangles an- to each other as the 
prodticts of their bases and altitudes. 

Ex. 1. Find the area ot a parallelogram whosa liuaa \s 'J tt. 8 ir.. 
and whose altitude is 2 ft. 3 in. 

Ex. 2. Piud th« altitudu of a trian^'lB wIiu^b iiica is 180 sq. hi. 
ana whose base is 1 ft. 3 in. 



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2u8 BOOK IV. WANE (iC(IMi:'r]:V 

PRoi'osiTioN VI. Theorem 

393. If a, h, c denote the sides of a tnuiigle opposite the 
angles A, B, C, respectively, and s = i {a + h -{- c) , the area 
of the triangle = V^s (s — a) (s — b) (s — c). 




Given the A ABC -withthe sides opposite A A, B and C, 
denoted by a, b and c, respectively, i (a + b -\- c) 
denoted by s, iiiid A an acute angle. 

To prove area A AB(7 = V^s {s — .i) (s-h) (.--^T 
Pioof. Draw the altitude BB and denote BB by h. 
Then a-^b- + c- — 2h X AB. Art. 349. 

.-. 2&X AB^b' + c^~<i:'. Axs. 2, 

....«.-±|^, 

But h'-<'' — A-D'-=(c + Al>) U — AD) Art. 3< 

=1"+^ — )\' — 2F^; •''■ 



. (i. + e + (0 (t + c — a) (»+!.-(!) (g — J + c) 



, a+ 6 — c = 2s — 2c, etc. 

Hyp,, Ass, 4, ; 



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AREAS OF POLYOONy '2?i^ 



4 6- 4 6- 



2\/,{s-a)U-b)U-c) 



B\it area A ABC^h b X h. Art- :m. 

•■. area AAiiC-v'i. (>,■ — «) (s — fi) {s~c). Ax, a. 
q. £. T>. 



Proposition VII. Theorem 

394. The area of a frapczotd is ei^nal In 
tm uf Hn hasen muUiplied by its altitude. 



Given the trapezoid ABCB with the bases AT) and BO 
(denoted by& and b'), and the altitude FB (denoted by h). 

To prove area of AB<7D=i {h + h') X h. 

Proof. Draw the diagonal HI). 

Then area of A AIW= h b X it. (Why t) 

And area of A BVD^i V X h. (Wliy?) 

Adding, area of AHCl) = h (W -^ b) h. (Why?) 

Q. E. D. 

395. Cor. The urea of a trapezoid equals the product of 
the ■iiD'd'dii of the trapezoid by the allitiide. For the njediaii of 
a trajjuKoid equals one-half the sum of the bases (Art. 179). 



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240 



,ANE GKO^CETKY 



S96. ycHOLiUM. The area of a polygon of four or more 
sides can usually be found in one of several ways; as, by 
dividing the polygon into triangles and lahhtg the sum of the. 
aiv.as of tkf. triangles; or, by draiving the loiige^tl diagonal 
of the polygon and drawing 
perpendiculars to this diago- 
•ual from the vertices which it 
does not meet, and obtaining . 
the smn of tlie- areas of the 
triangles and frupfzoids thus 
formed. 




e 5, 6, and 7 in. 
are IS and 10 iii.. 



Ex. 4. If tte area of a trapazoii] is 135, and i 
18, find its altitude. 



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COMPAKISON OP I'OLVGONS 



comparisom of polygons 

Proposition VIII. Theorem 

897. If two trianffles have an angle of one equal to an 
angle of the other, their areas are to each other as the pro- 
ducts of the sides includhuj the equal angles. 




Given the 
To prove 



, ABO and AJ)^' having Zi 
AABC__ABXAC 



in eommon. 



A ABF ADXAF 
Proof. Draw the line DC. 

Then the A ABC and ADC may be regarded &s having 
their bases in the line AU, and as having the uoinmon 
vertex C. 

A ABC A P. 
■-■AAIW^AJy ^''■'''- 



(Why t) 

l tlie corresponding members of these equali- 
A ABC ^ ABXAG 
A ABF ADX AF 



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IIOOK IV. PLANE GEOMETKY 



1'EOPUSITION IX. TlIEOHEM 

398. The arms of Iwu mnUar irkiitykn ar. 
as the squares of unij two hwiiuloijous shies. 





Given the similiii- & ABC and A'B'C with AB and A'B' 
homologous sides. 

A .4 lic _ ~n? 

To prove — -~", — r=zz:- 

^ AA'iJ'C" A'B'- 

Proof. Draw the homologous altitudes CF and C'F' . 

A ABC AB XCF AB ^ CF 



Then 



A A'B'O A'B'XC'F A'B' C'F' ' " 

arc to each other as the pymhicts of Ihfir hmes and aliitutlei)). 

CF A B 



Substituting TTTT/ foi' its equal 7^77^' 



A ABC ^ ^^ y^ --^J^  
A A'B'V A'B' A'B' 



AB- 
~VB'- 



Ex. ]. If a pair ot lioroologous siJes of two similar triangles 
4 ft. aud 5 ft., find the ratio of the areas of the triangles, 

Ex, 2. Prove Prop. IS Ijj use of Prop. YILI. 



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compjIeison of polygons 2i6 

Proposition X. Theohem 
399. The areas of tivo similar polygons are to each other 



s the squares of any two homologous sides. 




Given tiie similar polygons ABCDE and A'B'C'D'E', 
■with their areas denoted by S and S', respectively, and with 
AB and A'B' any pair of homologons sides. 

To prove S r S'=Xb^ : A'B''. 

Proof. Draw the diagonals AG, AD and A'C, A'B' from 
the homologons vertices A and A'. 

These diagonals will divide the polygons into similar & . 

Art. 32!l. 

A ABC AB^ 

. ^  Art. 308. 

_^A'B'G' A'B'"' 

" A^'B't" V7^^7 AA'Ci)' Vl^V 



AA'B'E' 
(Why?) 

A ABC ^ A AGB A ABB _ 
■'■ A A'B'V £\A'GB' AA'B'B' (^'^'y'^ 

A ABC + A AC I) + A ABB __ A AUG _ ^^^ .^^„ 
' A A'B'C'+ A A'G'B'-i- A A'B'E' A A'B'G' 

. '"^ ^ A ABC ^^ ,, 

•• «' A A'B'C" 

. ^ = J^. ^, 1 

^' -^I'-t;'' o. E. D. 



Ei. If a pair of honiologouH sides o! two site 
"at! 2 ft. , Cud the ratio of the areae o£ the polygoi 



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I'LAXE C.EOMKTKV 



Peoposition XI. Theorem 
400. Til a right triangle, ilie square on the htjpotenu! 
is eqtdvaknl to tht sum of the squares on the two legn. 




Given AT> the square on AC the hypotenuse of the rt. 
A ABC, and BF and BK the squares on the leys Ba 
;md BC respectively. 

To prove ADo BF + BK. 

Proof. Thi'oiigh B draw BL \\ AB, and meeting KB in 
B. Draw BE and FC. 

Then i ABC and ABG are rt. A . 

:. GBCls a straight line. 

Tn the A BAE and FAC, AB=^AF, i 



Also 



But 
(for AL hai 

Also 



ZBAB= IF AC, 
(for cach^ZBAC + arl. . 
:. A BAE=/\FAC. 



d AB=Aa. 

I. Why ?) 



In like 
Adding, 



square BF^2 A FAC. 
rectangle Ai— square BF. 
rectangle Ci~sqaare BK. 
AL + CL, or AB^BF+BK. 



As. : 



J A BAE). 
(Why ?) 



401. Cor. 2'ke square on either leg of a right triangle 
is eqiiivalent to the nquare on the hypotenuse ditninished by 
the square on the other leg. 



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CONSTEUCTION I'iiOBLEMB 



COKSTKUCTIOK PROBLEMS 



Proposition XII. Problem 

402. To construe/, a square equivalent to the sum of iwo 
given squnrex. 



c 




A Ir 

Given two squares S and S'. 
To construct a square oquiviilciit to S 4- •'^'-  
Construction. Constrnct a right angle BAC. ah. 574. 
On one side of this angle -take AB equal to a side of S, 
and on the other side take AG equal to a side of S'. 
Draw BC. 

On a line equal to BC construct the square B. 
Then B is the square required. 

Proof. B=BC^'^'Jb'^ + A(f. Arf.4r)0. 

.-. B=^S+ iS" Ax. 8. 

Q. E. F. 

403. Cor. To construct a sqtiare e<inivule}i.l to the sum 
of three or more ghm squares. At C in the above figure 
erect a liue GJ) X BC (Art. 274), and equal to a side of 
the thii-d given square. Draw DB. DB will be a side of 
a square equivalent to the sum of three given squai'cs, etc. 



Ez- 1. CoBstruot a square equivalent to ths s 
■whose sides are i in. and 1 in., respectively. 

Ex. 2. By use of Art. 40J, taking a given Um 
v'S; also v/'l. For example, eonstraet 



1 o( t 



) squares 



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^'■iO liOOK lY. PLANE GKOMETET 

PiiOmsinoN Xin. 3'roblkm 

404. Tomm'InicI a ^qiuirc equivalent (o the difference 
of two yiven aquare^i. 



B 



B 



Given the squares W aiul S'. 

To construct n sqnarij equivi 
and S'. 

Construction. Coustn 



hnit t.o iln; difTnrcncR of ; 



■t'Ait&ngh BAR'. 



Art. 274, 

Ou one siilo J B take AB equal to a side of the smaller 
given square 8'. 

From B as a center with a radius BC, etjnal to a side of 
the larger square, describe an are intersecting AK in G. 

On a line eqnal to AC eoustruct the square R . 

Then B is the square required. 



Proof. 



B = AC-^BC- — AB', 



Art, 401. 

{the nqiiare on either leg of a right triangle is equitiateni to the square on 
the hypotenuse diminished by the square on the other leg). 



Ex. Construct fl square e 
whose sidtB are 1 in. and i ii 



Q. E. F, 
ilitTereuce of two aquarea 



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CONSTRUCTION PROBLEMS ^H 

Proposition XIV. Problem 
405. To cimstriid a square eqxdmhnt to a given paml- 
lelogram. 



i) k -p  

Given tlie CJ AliCT) with base S and altitude h. 
To construct a sciuave equivalent to ABCD. 

CoEstructioa, On the lino EO take EF (sqnal to h and 
FG equal to 6. 
On EG iis a diameter construct a semicircle. Art. 27a, Post. 3. 

At F erect a X meeting the semieircumference at K. 

Art. 274. 
On a line equal to FK construct the square S. 

Then S is the required square. 

Proof. S^KF'. 

But KF''^bXh, Art. S43. 

(n ± from a«ij poiii! hi a cinyiimfcrnnw lo a i^iaJiii^lci- is n mean pyopor- 
tioiial iftwecn lite sajotenls of the diameter). 

But area CJ ABGD^b X h. (Why t) 

.-. 6'oarea £Z7 ABGD. 

Q. E. F. 

406. Cor. To construct a square equivalent to a giien 
triangle, eonstriiet the mean proportional between the base 
, and half the altitude of the triangle and construct a square 
oa ihis mean proportional. 



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i8 BOOK IV. PLANE GEOMETRY 

Peoposition XV. Problem 
407. To construct a triangle cqmvaleid to a give 




Given the polygon ABCBE. 

To construct a triangle equivalent to ABCDE. 

Construction. Let --i, B, (7 be iiny three consecutive ver- 
tices in the given polygon. 

Draw the diagonal AC. 

Draw BF \\ AC (Art. 279), and meeting AE produced 
at#. 

Draw FC. 

In the polygon FGDE take the three consecutive ver- 
tiees C, B, E, and draw the diagonal CE. 

Draw BG || CE, and meeting AE produced at G, Draw 
CG. 

Then A FCG is the triangle required. 

Proof. A ABC->AAFG, Art. wo. 

(having the same base AC, and their veriiees in a live BF |] the base) . 

Also A AGE = A ACE. Went. 

And A FGB = A ECG, Art. 390. 

[hamng the same base CEand their vertices in line DG [j >Mse). 

Adding, A ABC + A AGE + A ECB o A AFC + 
A.1CB+ A ECO. As- 2. 

Or polygon ABCDE<^ A FCG. 

»}. ffi. F. 

408. Cor. To construct a square equivnUnt to a gi^en 
polygon, use Arts. 407 and 406. 

Ei. Construct a triangle equivalent to a given hesagon. 



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CONSTRUCTIOK PROBLEMS M\) 

Proposition XVI. Pkoblem 

409. To construct a rectangle equivalent to a given square, 
and having the sum of its iase and altitude equal to a given 
line. 



r D. 



Given the sf|uiire S and the line ,4 B. 

To construct a rectangle equivalent to S, nnd having the 
sum of its base and altitude equal to AB. 

Construction, On AB as a diameter describe the semi- 
circumference ADB. Art, 275, I'OBt. 3. 

At the point A erect a X , AG, equal to a side of S. 

Art. 274. 

Through C draw a line || AB (Art. 279), and meeting 
the cLryumference at I). 

DrawDE J. AB, Art, 273. 

Construct the rectangle R with a base equal to EB and 
an altitude equal to AE. 

Then B is the rectangle required. 



Proof. 


DB- = AJ'JXEB. 


(Why f) 


But 


DE=CA. 


(Wby ?) 




.: CA^=^AEXEB. 


(Why?) 


Or 


S=>B. 





410. Cor. The above problem is eqiiivaieiit to the 
problem: Given the sum and product of iivo lines, to con- 
struct the lines. 



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50 BOOK IV. I'LANE GIilOMETKY 

Proposition XVII. PiioJiLEJi 

411. To eon&iruct a rectangle equivalent to a given square, 
'tid having ike difference of Us base and altitude equal to a 
liceu line. 



[ZZZl 



Given the square S and the line Ali. 



To construct a 
difference of its 1: 



ictangle equivalent to .S', and 
ie and altitude equ<il to AB. 



Construction. On AB as a diameter describe the c 
fereuce ABBF. Art. 275, Post. 3. 

At A erect the X AC equal to a side of S. Art. 274. 

Draw C'F through the center 0, and meeting the circum- 
ference at the points D and F. 

Construct a rectangle E with base equal to CF and alti- 
tude equal to CD. 

Tht:ii li is the rectangle required. 

Proof. 



CF: CA = CA : CB, 


Art. -isa. 


.: CA'=CFX CIK 


(Why !) 


:. S^B. 


(Why ?) 



Also the difference of the base and altitude of B= 
CF—CB = DF=AB. 

Q. E. F. . 

412. Cor. The above problem is equivalent to tbo 
problem: Given ike difference and the product of two lines. 
to constrtKt the Ums, 



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CONSTRUCTION PROBLEMS ^Ol 

Proposition XVIII. Problem 
413. To construct a polygon similar to two given simi- 
lar polygons, and equivalent to their sum. 




B' K"""L 

Given the similar polygons P and P'. 

To construct a polygon similar to P and P', and equiva- 
lent to their sum. 

Construction. Take any two homologous sides, AB and 
A!B', of P and P'. 

Draw J/A'X KL (Art. 274). making l/i:=J.B, and ^i 
= A'B'. 

Draw ML. 

On A"B", equal to ML, as a side homologous to AB 
construct the polygon P" similar to P. An, 3T2. 

Then P" is the polygon required. 



Proof. 




I" A«B«'' "'■" P" ^"B"'' 




Art. 399. 


(a. „r,a, 


o/e 


'1-rt similar polygons ore k> mch other a. 
their homologous siiles). 

, P+F' lYf + A^'^ 


? Ihe 


*9"a'"i^» 0/ 


Atldiii?. 






As. a. 






^P" ^ .1"P"^ 






But 




jU"2i:^"+ A'i^ = i¥//, 




Art, 346. 


Or 




AB--\- A'B'- 'a"B"^_i 




As. 8. 
As. 8. 



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BOOK IV. I'LASE GEOMETKY 



Proposition XIX. Proulem 

414. To construct a square which shall have a given 
ratio (o a given square. 



B 



A'-^:[yW"";c 



Given the square S and the lines m a.nd n. 

To construct a square which shall be to S In tlie ratio 



Coastruction. Take AB equal to a side of S and draw 
AF, making a eonveuieut angle with AB. 

On AF take AD equal to m, and J>F equal to w. 

Draw DB. Draw F€ \\ DB, meeting AB produced in C. 
Art. 270. 

On AC, as a diameter, construct a semicircumference 
AKC. Art, 27-1, Post. 3. 

At B erect si L BK meeting the semieireumference at Tf. 
Art. 274. 

Construct a square B' having a side equal to BK, or x. 

Then &' is the square required. 

Proof. ^^ = a X 6. (Why?) 

Also a : h = m -. n. (Why?) 

□„„„„ «_«!_«!_"_'». ,,. . . 



S' 



ab b 



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OOKSTRUOTION PRORLEMa 2:)d 

Proposition XX . Pboelem 

415. To construct a polygon similar to a given polyg&ii, 
ind having a given ratio to it. 



Given the polygon 7" and the lines m and n. 

To construct a polygon /-" which shuU be similar to P, 
and be to P in the ratio n : m. 

Construction. Construct a square which shu.!! be to clie 
square on AB as n : ni. Art. m. 

Let A'B' be a side of this square. 

Upon A'B' as a side homologous to AB construct a 
polygon P' similar to P. Art. 37'J. 

Then 1" is the polygon required. 



p_ 


AK 


I' 


375 


AJP 


-". 


JTb'- 


n 



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PLANK GEOMETIIY 



Pkoposition XXI. Problem 

416. To construct a polygon similar to one given polygon ^ 
and iquiviiUnt to another given polygon. 



O^ 



Givea the polygons P and Q. 

To construct a polygon similar to P, and equivalent to Q. 

Construction. Conatruet a square equivalent to P, and 
let in tie one of its sides. Act. 408. 

Construct a square equivalent to Q, and let n be one of 
its sides. Art. 4oa 

Construct >1'/J', the fourth proportional to ni, n, and AB 

Art. 3GG 

On vl'-B', as a side homologous to ^l/>', euiistruet a poly 

gon P" similar to P. Art. a72, 



Then P is the polygon required. 






^ , P m^ Tli^ P 
Proof. _= =.^^=-_. 


Constr 


, Arte. 314,399. 


P P 




As. 1. 



.-. F^Q. 



Art. 305. 
Q. I. F. 



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EXERCISES . THEOREMS 2 JO 

EXERCISES. CROUP 4«> 

THEOEF.>[S CONCERNING AREAS 

Ei. 1. The dia{;oiial3 of a parallelogram divide the parallelogram' 
into four equivalent tnangles, 

Ex. 2. Any straight line drawn through the point o£ intersection 
of the diagonals of u parallelogram divides the parallelogram into two 
equivalent parts. 

Ex, 3. If, in the triangle ABC, D aud F are the midpoints of the 
sides AJi and AC, respeetively, the area of ADF equals one-fourth the 
area of ABC. 

[SuQ, Use Art. 397,] 

Ex. 4. If the midpoints of two adja.'ent aides of a pruallologram 
be joined, the area of the triangle so fonned equals oiii^-eighth the 
area of the parallelogram. 

Ex. 5. If, in the triangle ABC, I) and /■■ are the midpoints of the 
sides AB and AC, respectively, the triangles ADC and AFB are equi- 
valent, 

Ex. 6. In a right triangle show, by obtaining expressions for the 
area of the figure, that the product of the legs equals the product of 
the hypotenuse by the altitude upon the hypotenuse. 

Ex. 7. If two triangles are equivalent, and the altitude of one is 
three times the altitude of the other, find the ratio of their bases. 

Ex 8. If two isosceles triangles have their legs equal, and half 
of the base of one equivalent to the altitude of the other, the tri- 
angles are equivalent. 

Ex. 9 If two triangles have an angle of one the supplement of 
an angle of the other, their areas are to each other as the products of 
the sides ineludinj; these angles. 

(I b 

Ex. 10. Prove geomiitrieally that (ii+ii) =" •!■'■ [ ~~1~ 
+2(ib. 

Ex, 11. Similarly prove («—i)'=a' + i''~2u6. 

Ex 12. Similarly prove (o+f)}(n—;>)=(i=—i/'. 

Ex. 13. The line joining the midpoints of the parallel aid^s of u 
trapezoid divides the tvupezoid into two i^iiuivitliini piirt^.. 



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ii56 



"BOOK IV. TLANr; GKOMETUV 



Ex. 14. The linea joiuing tlio miiipolTit of otio diagonal ot i 
quadrilateral to the verticea not joined by 

the diagonal divide the quadrila,teral into B T O 

two equivalent parts. / / 

Ex. IB. Given QR and TS passins 
through P, any point on the diagonal AC 
Df a CJ , QR\\ AD, and TS ]] All, prove 
QBTP<:^PRDS. 

Ex. 16. Given OH^OD; proveAABC^ 
AADC. I-et the pupif also state this as a 



USE Of ACXLLIARY LINES 

Ex, 1. Given AJICI) a ZZ7 and P any 
point inside ABDC; prove APJD + APBG 
= £:.PAB+APCD. 



Ex, 2, The area of a triangle is equal to 
one halt the product o£ its perimeter by the radius of the inscribed 

Ex. 3 If the estremitiea of one leg of a trapenoid bo joined to the 
midpoint of the other leg, the middle one of the three triangles thus 
formed is equivalent to half the trapezoid. 

Ex. 4. Tho area«f a trapezoid is equal to the product of one leg by 
thn pi'rpeudicular on that leg from the midpoint of the other leg. 

Ex 5. If the midpoints of the aides of a quadri- 
lateral be joined in order, the parailelogiam thus 
formed is equivalent to one-half the quadrilatetal 

Ex. 6. A quadrilateral is equivalent to i triangle 
two ot whose aides are the diagonals of the q u id i: lateral, the angle 
included by these sides being equal to one of the angles formed by 
the intersection of the diagonals. 



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EXERCISES. THEOREMS J,i>i 

EXERCISES. CROUP 4S 

THEOEEMS PROVED BY VARIOUS METHODS 

Ex. 1. If through the midpoint of one leg of a trapeioid a line be 
drawn parallel to tde other leg to meet one base and the other base 
produced, the parallelogram so formed is equivalent to the trapezoid, 
Ex. 2, If the midpoints of two sides of a triangle be joined to any 
point in the base, tbo quadrilateral so formed is eqaivalent to halt the 
triangle. 

Ex. 3. If P is any point on AC the diagonal of a paralleloeram 
ABCD, the triangles AFB and APD are equivalent. 

Ex. 4. If the siili) of an equilateral triangle be denoted by a, tha 
area of the triangle equals -~r~- 

Ei. 5. Find the ratio of the areas of two equilateral triangles, i£ 
the altitude of one equals the side of the other. 

Ex. 6. If perpendiculars be drawn from any point within an 
equilateral triangle to the three sides, their sum is equal to the alti- 
tude of the triangle. 

Ex. 7. If -E is the intersection of the diagonals AC and SD of a 
quadrilateral, and the triangle ADE is equivalent to the A £EC, then 
the liues A IS and CD are parallel. 

Ex, 8. If, in the quadrilateral AliCD, the triangles AUC and AIiC 
are equivalent, the diagonal AC bisects the diagonal BD. 

Ex. 9. If two triangles have two sides of one equal to two sides 
of the other, and the included angles supplementary, the triangles are 
e(iuivalent. 

Ex. 10. II, in the parallelogram -4lJ?CD, 
F is the midpoint of the side BC, and AF 
intersects BD in K, the triangle liEF=-f^ the 
parallelogram ABCb. 

Ex. 11. Given PQ [[ AC, and PR \\ AB; ^ 

prove AQAB a mean proportional between AHQP and AFRC. 

Ex. 12. P Ik any point in the side llC 
of liie parallelogram A/!(-lr and HP pro- 
duced meets .1 /( produced iu <l Show that 
the triangles BPA and CPQ are equivalent. 




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258 BOOK IV. PLANE GKOMRTIiY 

EXERCISES. GROUP 43 

PROliLEMS IN CONSTRUCTING AREAS 

Ex. 1. Construct ii square having twii;*; Ihi; nrea i)£ n given square. 

Ex. 2. Construct a square hiiviug three times tho area of a given, 
square. 

Ex. 3. Construct a square oquivaient to tbe sum of three pivett 
squares. 

Ex. 4. Transform a given triangle info an equivalent isosceles 
triangle having the same base. 

Ex. 5. Transform a given triangle into an equivalent triangle 
having the same base, but having a given angle atjjaeeut to the base. 

Ex. 6, Transform a triangle into an equivalent triangle with the 
same base, but naving another given side. 

Ex. 7. Transform a parallelogram into an equivalent parallelo- 
gram having the same base, but ooutaining a given angle. 

Ex. S, Construct a triangle similar to a given triangle ami con- 
taining twice the area. 

To construct a similar triangle containing five times the area: how 
is the construction changed ! 

Ex 9. Bisect a given triangle by a line parallel to the b^se. 

Ex. 10. Constmet a polygon similar to tivo given i^imilar pohgons, 
and equivalent to their difierence. 



Ex.11. Draw a line parallel to one 


side of a given rectangle, and 


cutting off five-sevenths of the area. 




Ex.12. Bisect a parallelogram by a 


lino perpendicular to the base. 


Ex. 13. Through any given point c 


km , line U„Mme , t-iyea 



parallelogram. 

Ex. a. Construct a triangle equivalent to 
a given triangle, ABC, having a given base 
AD, but the ZBAG adjacent to the base un- 
changed. 

[SuQ, Draw CE \\ DU, etc.] 



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EXEHCISES. PEOBLEMS 259 

B:i. 15. Transtorm a parftllelogram into an equivalent parallelo- 
giaxa haviagagivenbaee, but the angle adjacent to the base unchanged. 

Ex. 16. Tranaforra a given triangle into an equivalent riglit 
triangle having a given leg. 

[SUO, Uhb Ex. 14, then Ex. 5,] 

Ex, 17. Transform a given triangle into an equivalent right tri- 
angle having a given hypotenuse. 

[SuG. Piod the altitude upon tlie hypotenuse of the new triangle 
by fijsding the fourth propovtionnl to what three liiu's ? ] 

Ex. 18. Transform are-entrant pentagon into ; 
equivalent triangle. 

Ex. 19. TranBform a given tnaugle ioto 
equivalent equilateral triangle. 

[Sue. See Art. 416.] 

Ex. 20. Biseat a triangle by a line perpendicular to one of its aides. 

[Sue. See Art. 416,] 

Ex. 21. Construct a square equivalfiit to two-thivila of a given 
square . 

EXERCISES. CROUP 44 

PROBLEMS SOLVED BY AEUEBHAIC ANALYSIS 
Ex. ]. Transform a given rectangle into an equivalent rectangle 



with a given base. 




Ex. 2. Transform a given square i; 
given leg. 

Ex. 3. TrauBform a given triangle 
right triangle. 


ato a right triangle having a 
into an equivalent iaosneiea 


Ex. i. Draw a line cutting off from 
triangle equivalent to one-half the give 
[SuG. Use Art. 3SIT.] 


, a given triangle an isosceles 
a triangle. 


Ex. 5. Through a given point in 
draw a line biaecting the area of the tri 


one side of a given tz'iangle 
angle. 



Ex. 6. Transform a giveu square into a rectangle which shall have 
three timeu th« jietimetec oi the given square. 



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Zbi) BOOK IV. PLANE GF.OMETHY 

EXERCISES. CROUP 4S 

PR0RLE3IS S0LVI3D BY VARIOrS >[ETHODS 

Ex, 1, Bisect a given, paralle log vara by a line parallel to the base. 

El. 2. Tmnsforni a pavallBlogr.am into an equivalent parallelo- 
gram having the aame base and a given side adjacent to the base. 

Ex. 3. Construct a square which shall aontain four-seventhg of 



the area of a given square. 










Ex.4. 


In two different ways 


i constr 


uct a square t 


.aving 


three 


times the 


areaof aglvensfjuare. 










Ex.6. 


Trisect a given triangl 


e by line 


IS piiraliel to thi 


i base. 




Ex. 6. 


Find a point ivithin a 


triangle 


such that line; 


i drawn 


from 


it to the \ 


'erticestriseet the area. 










Ex. 7. 


Find a point within a 


triangle 


such that iinci 


•. drawn 


from 


it to tha 


three vertices divide the area i 


iito parts whic 


■h shall have 


the ratio 1 


2:3:4. 










[Sue. 


II one of the small A 


eontains 


i tlie area o£ . 


original 


A, a 


line thcoi 


igh its vertex cuts off s 


the altitude, etc.] 






Ex. 8. 


Divide a triangle ini 


to tbree 


equivalent parts by 


lines 



Ex. 9. Divide a triangle into three equivalent parts by lines drawn 
through a given point J' in one o£ the sides. 

[Suo. Use Art. 307.] 

Ex. 10. Divide a giveu qiiadrilatei-al into three equivalent parts 
by lines drawn through a given vertes. 

Ex. 11. Through a given point in tlie G 

base of atrapezoid draw a linebiseeting the ,' ''., 

area of the trapezoid. / \ 

Ex. 12. Bisect the area of a trapezoid y ' \ p 

by a line drawn parallel to the bases, p[ \ p 

[SUG. Construct A GKF similar to A J \ p 
ABG aud equivalent to i num el wliat two 



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Book V 

REGULAE POLYGONS. MEASUREMENT OF 
THE CIRCLE 

417, Def, a regular polygon is a polygon tliat U both 
equihiteral aod equiangular. 

Proposition 1 . Theoreie 

418. An equilateral poiyijnii (hat in in'^i-rihed in o circle 
is ulso eqaiaiigtilar and regular. 




Given ABC . . .K an inserilied polj^goii, with its sides 
at;, nc, CD, etc., etiiia!. 

To prove the polygon ABC...K equiangular and 
rogular. 

Proof. Arc A7.' = ai-c mj^avc- Cl>, ptc. Art. 21M. 

.-. are ABG^ayq. BCI}=arc CJ>E, etc. Ax. 2. 

.-. ^ ABC^IBCI)^ ^CDE, etc.. Art. 2C0. 

(all 1 inscribed in the same segment^ or in equal segments, are equal). 

.: tlie polygon ABC , . . Jt is equiangiiki'. 

/. the polygon ABC . . . KU regular, Art, 417. 
0, z. B. 

{■JCI) 



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262 



KOOK V. PLANE GEOMF.TKY 



419. Cor. 1. // the arcs suhtended by the sides of a 
regular insrrih''(l polygon he bisected, and each point of 
bisection lie joined to the nearest vertices of the polygon, a 
regular inscribed polygon of double the number of sides is 
formed. 

420. Cob. 3. The perimeter of an inscribed polygon is 
less than the perimeter of an inscribed polygon of double ike 
number of sides. 



Proposition II. Tiif-orem 

421. A circle may be circumscribed about, and a circle 
may be inscribed in, any regular polygon. 




Given the regular polygon ABCIIE. 

To prove that a O may be circi-imBeribeil about, or in- 
scribed in, ABCBE. 

Proof. I. Throueh A, B aad C (Fig. 1) , any three suc- 
cessive vertices of the polygon ABCDE, pass a circum- 
ference. Art. 235. 

Let be the center of this eircumferenee. 

Draw the radii OA, OB, 00. Also draw the line OD. 

Then, in A OBG, OB=OG. (Whj-r) 

:. lOBC^lOCB. (Whjf). 



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KEGULAE POLYGONS 263 

But Z ABG = Z BCD, Art. 417. 

(being A of a regular j)Olii3on). 
Subtracting, Z OBA = Z 00i>. (Why?) 

Hence, iu the A OAli and OCD, 

OB=OG. (Whyt) 

^B-Oi). (Why?) 

Z OBA = Z OC'D, 

(JH;;! jtniuecf). 

.-. A ABO^AOVB. (Why!) 

.-. 01>=0A. (Why?) 

Hence the cireiirafereuce whieli passes through the 
vertices A, B and C, will iilso pass throuirh the viirtes I). 

In like manner, it may bo proved that this circumference 
wdl pass tbrongh the vertex E. 

Hence a circle described with as a center, and OA as 
a radius, will be circumscribed about the given polygon. 

H. The sides of the polygon ABODE (Fig. 2) are equal 
chords iu the circle 0. 

Hence they are equidistant from the center. Art. 22(i. 

.■. a circle described with as a center, and the distance 

from to one of the aides of the polygon as a radius, will 

he inscribed in the given polygon. 

q. E. B. 

422. Def. The center of a regular polygon is the com- 
mon center of the inscribed and circumscribed circles, as 
the point in the above figure, 

423. Def, The radius of a regular polygon is the radius 
of the circumscribed circle, aa OA in the above figure. 

424. T>FF. The apothem of a regular polygon is the 

radius uf tliL- inscribed circle. 



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2fJ4 BOOK V. PLANE GEOMETRY 

425. Def. The angle at the center of a regular polygon 
is the angle between two radii drawn to tlie extremities of 
any side, as the augle AOB. 

426. COE. The angle id the. center of a re</Hlar polygon 
is equal to four right angles divided l»j the ntiinher of sides. 

Hence, if n denote the number of sides iu the polygon, 

the angle at the center of a regular jiolygon eijuulx —~ — ; 

also Ike angle between un apotkem and the nearest radius, 

in a regular polyjoit of ii sidvs, eqiiaU   

PHO^osITIO^f III. Theorem 

427. If the circumference of a circle be dirided info any 
number of equal ares, 

I. The chords of these arcs form a regular inscribed 
polygon; 

II. Tangents to the circrimference at the points of di>:ision 
form a regtdar circumscribed polygon . 



e^:^c 



Given the circumference ABC, divided into the equal 
arcs AB, BC, GJ), etc., the chorda AB, BC. etc., and PQ, 
QB, etc., lines tangent to the circle at B, (J, etc. 

To prove ABODE a regular insei-ibcd polygon, and 
PQRST a regular circumscribed polygon. 



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eegUlak polygons Libu 

Proof. I. The chords AB, BC, CD, etc., are equal. 

(Why r) 
.■- po'ygoii ABODE is equilateral and regular. Art. 418. 

II. Ill the A APH, BQG, CRD, etc., 

AB=liC=Cn, etc. (-Why!) 

Also ZP.4B- ZPfiA = IQIiC^ IQGB= ^ BCD, etc. 

Art. 264. 

{«id< la:. 

.'. A AZ-'7i, /iX'<-''. Clilt, lite,., areeqiiiil, isosceles triangles. 

(Why?) 
.-. ^1'-^ IQ ^ Z7i*, etc. (Why?) 

Afld AP^PB^BQ=QC, etc. (WhyT) 

.-. PQ=QB = nS, etc. Ax. 4. 

.-. PQBST is a regular i)Cilygoa. Art. 417. 

Q. E. ». 

428. Cor. 1. If the arcs AB, BC, CB, etc., be bisected, 
and (1 tangent fie draimi at each point of bisection , a circum- 
scribed regular polygon of double the number of sides of 
PQI18T will be formed. 

429. Cor. 2. The perimeter of a circumscribed regvlar 
polygon is greater than that of a circumscribed regular poly- 
gon of double the number of sides. 



Ex, 1. Fiud the number o!^ <Jep:vees lu the central angle of a 
legalar poiitagon. Of a regular hexagou. Of a square. 

Ex. 2. WUat is the sliovt name for an iuscribeii equilateral quad- 
Tilu^al ; 



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26G BOOK V. PLAVE GEOMETRY 

Proposition IV. Theorem 
430. Tangents to a circle <at the midpabits of the arc? 
subtended hij the sides of a regular inscrihed pobj'jon form a 
regular circumscribed polygon whose sides U7'€ purulUl to 
the corresponding sides of the inscribed polygon. 




Given the regular polygon ABODE iiiscribed in tlie O 
ACD; P, Q, B, etc., tiie midpoints o£ tiie arcs AB, BO, 
CJ), etc.; and A'B', B'G', CD', etc., tangents to the circle 
at P, g, E, etc. 

To prove A'B'C'D'E' a regular polygon with its sides || 
corresponding sides of the polygon ABCDB. 

Prool The ares AB, BC, CD, etc., are equal. Ait. 218, 

.-. the arcs AF, FB, BQ, QG, etc., are equal. Ax. 5. 

.■, the arcs I'Q, Qli, BS, etc., are equal. Ax. 4. 

.■. A'B'C'D'E' is a regular polygon, Art. 427. 

{if Ike cii-ciimfereiiee of a O ho diiikkd, clc.). 

Side AB A OP. Art. us. 

A'B' ± X>P. (Whyt) 

.-. AB II A'B'. (Why?) 

In like manner, each pair of homologous sides in the 

two polvgons is parallel. 

Q. E. D. 

431. Cob. Homologous radii of an inscribed and a eir- 
eumscribed regular polygon, wlhse sides are parallel, coin- 
cide in direction. Thus, in the above figure A POA and 

';. 42e° 



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EEGnLAB POLYGONS 



Proposition V. Theorem 



432. Two regular polygons of the same number of sides 
re similar. 




Given K and K' two regular poJyf>ons, eaoli of n sides. 

To prove if and K' similar. 

Proof. Eaoh Z of A'^ -■■ ''  '" " -■-- -" ■■- •  - -. Art, 174 

(i» an equiangular pohjgon ff n sides, each Z = — - — - — )  

Similarly each Z of IC—   "  "  '  — 

Hence K and K' are mutually equiaog'ular. As, i, Art. igs 
AB_ 



Also 


AB = BC .-. fi;,.l. , 


Art. 417, Ax, ii. 


And 


a-b:b'<7 ,. ff;.l. 


(Why!) 




. AB^A'B' ^|, AB _ BO 
"" fiC B'C' A'B' B'C' 


(Why ?) 


In like 


BO CD BE ^ 
manner ^gJ^^-p^-J^,, elo. 




Henee K and K' have tlieir homologous 


sides propor- 


Hence K and K' are similar. 


AH. 221. 
Q. E. D. 



433. Cor. The areas of lico regular polygons of the 
same mmibe.r of sides are to each other as the squares of any 
two homoloyoHs sides. 



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2G8 BOOK V. TLANE GEOMETRY 

Pkopositiox VI. Theokeji 

434. I. The perimeters of ttco regular polygons of the 
same numher of sides are to each other as the radii of their 
circumscribed circles, or as the radii of their inscribed 
circles; 

11. Their areas are to each other as the squares of these 
radii. 




Given AC and A'C two regular polygous, each of n 
sides, with centers and C, and with perimeters denoted 
by P and J", radii by E and R', and apothems by r and »■', 
respectively. 

To prove. I. P : F' = R -. E' = r -. r'. 

II. Area AO -. area A'C' = R^ : R'-^r- -. r'-. 
Proof. I. Thepolygons ACand A'Care similar. Art. 432. 
Hence P : P-^AB -. A'B'. Art. 34T. 

But, in the A OAB and (yA'B', 

I AOB = lA'O'B', [foreai-h Z=l^). Art, 42G. 
Also OA : OB^O'A' : O'B', (for each A is isosceles). 

.-. A OAB and O'A'B' are similar. Art. 327. 

.-. AB : A'B'=OA : CfA'. Art. 321. 

And AB : A'B'=OL i O'L'. Art. 338. 

.-. P : F^OA : O'A'^OL : O'L'. Ak. i. 

Or F: P'^B: R'^r: r'. 

II. Area AC : area A'C'='AB^ -. A'B'^. Art. 399. 

But AB^ ■.A^'- = R' : R'" = r^ : r'-. Art. 314. 

.'. area AG ; area A'C'^B- : £'• ; r : r'^. As, l. 



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EEGULAE POLyGONS ZW 

Proposition VII. Theorem 

435. If the numher of sides of a regular inscribed poly- 
gon he indefinitely increased, the apothem of the polygon 
approaches the radius as a limit. 




Given the regular inscribed polygon AB . . . D oi n sides, 
with radius OA and apothem OL. 

To prove that, as n is indeiinituly Ln(;reascd, OL ap- 
proaches OA as a limit. 

Proof. In the A OAL, AL> OA — OL, Art. 93. 

{any siite of a ^ is greater than the lUffcrencn hchrccii the 
other two sides.) 

But, asn^^,AB = .: AL^O. Art, 25a, 3. 

Hence OA—OL = 0. 

:. OL = OA, or )- = JJ. 

Oi' the limit of thf apotliera OL is the nidius OA. 

Q. s. J>. 

436. Cor. As« = ^, iJ^— f- = 0. 

For, iS^ — »-2={K + r) (iJ — r). But, as »(=», 

B+r^R + E or 2 R, and [i — r = 0. As, s. 

.-. S2^,-- = 2i," X 01- 0. Ai<. s. 

Ex. A pjiir of bomologous sidfis of two regiiiai' pentagans are 2 
and 3 ft. Find ttn) ratio of tlie areaa of llio yglygons. 



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liOOK V. PLANE GEOMETRY 



Proposition VIII. Theorem 

437. The hnffHi of any line incloning Ihe circrim/crence 
of II circle, and Jtot passing within the circumference, is greater 
than tlie length of the circumference. 




Given the cirenmferenee AKEF, and ABCDEF any line 
which does not pass within AKEF. 

To prove cireumferenee AKEF < perimeter ABCDEF. 

Proof. Let ST be any point on the circumference AKEF 
not touched by the line ACBEF. 

At K draw a tangent to the given circle, meeting ACEE 
at B and I>. 

Then the straight line BKB < line BCD. Art. 15. 

To each of these unequals add the line IlEFAB. 

Then line ABKDEF < line ACBEF. Ax. 9. 

Henee every, line enveloping the circular area AKEF, 
except the circumference AKEF, may be shortened. 

Hence the circamference AKEF is shorter than any line 

enveloping it. 

'^ " Q, S. B., 



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REGULAR POLYGONS 271 

438. Cor. 1. The circumfef&iice of a circle is less than 
the perimeter of any polygon circumscribed about the circle. 

4.89. Cor, 2. The circumference of a circle is greater 
than the perimeter of any polygon inscribed in the circle, 
for each side of an, inscribed polygou is less than the aro 
subtended by it. 

440. Cor. 3, The difference between (he perimeters of 
an inscribed and a circumscribed polygon is greater than the 
difference between either perimeter and the circumference of 
the circle. 



Proposition IX. Theorem 

441. If the number of sides of a regular inscribed, or 
of a regular circumscribed polygon be indefinitely increased, 

I. The perimeter of each polygon approaches the circum- 
ference as a limit; 

II. The area of each polygon approaches the area of the 
circle as a limit. 




Given a circle of eircnmfei-ence (7 and area A, with regu- 
lar inscribed and cireumscribed polygons, each of n sides, 
with their periraetei's denoted by P and P', and tijeir areas 
by AT and K', respectively. 

To prove that as n is indefinitely increased, P and P' 



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2i2 BOOK V. PLAXE c.T-,o^iEn;Y 

each approaches C as a limit,, au,l K and K' each ap- 
proaches A as a limit. 

Proof I. Denote the apothems of tlie two polygons by 
H and )■. 

Then j,=''.- Art-™- 

Hence ^^ =^^- '^''- ^^"■ 

Let 11 = 0=; then Ji—r = 0. Art. 43ii. 



Rut p'— < f"— 7' (Art. 440) .-. 1"—C^0, or F = C. 
Also C— P < i"— .?' (Art, 440) .-. C — P=n, or P^C. 

K 

Let H =1. :o ; then 7^- - (■= i 0. An. 43G. 

... ^^'" "T . ..ll i 0. Art- 253, 4. 

(/(J)' Wic d(noinin'i(or , J'', infciscs). 
H^nee — J=-^ i (As S), Z. K' ~ K= 0, (Art. 253, i). 

But ,ff'— ^1 < iL'— .ff(Ax. 7) .-. .B:'— A=0, or ff' = J.. 
Al::;o A — K<K'—K{k:^.r) :. A — K=0, or K^A. 

Q. E. D. 

Ex, Find the perimeter and firea of a square field, one of whose 
BldBS is 10 rods. Find the same in a square iield, oni: of wlioso sides 
is 20 rids. Is it more economical, thei'eforf, to leuee land in largs 
oc small gelds t 



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REGULAR POLYGONS 



Proposition X. Theorem 
442. Two circumferences have the same ratio as (kei 
radii, or as thHr dia 




Given the circles and (7, with eireuniferenees denoted 
by Caiid C , mdii by R and }}', diameters by D and i/, re- 
Kpoci.ively. 

To prove C : C = R : R' = D : D'. 

Proof. Ill the given ® let regular polygons of the same 
number of sides be inscribed. Denote the perimeters of the 
inscribed polygons by P and P', respectively. 



Then 



F' It' 



lience, by alternation, JT^jv,' ■A"''- 30'/. 

If, now, the number of sides of the similar inscribed poly- 
gons be increased indefinitely, P becomes a variable ap- 
proaching C as a limit. Art. 441. 

P 

Hence— becomes a variable approaching — as a liri 



' E 



' Ji" 



But the v; 


riable 


— =the variable 






C 


c 






 R 


' R' 






:. c : e- 


-R: R 


Also %-,-- 


VL 


Art. 315.) 


(' 



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2 1 4 ROOK y. I'LANK flKOMiyi'KY 

rROPOSITlON XI. TnEOIlKM 

443. In Cirri/ circle, the ratio of the circumfcrenre to Ikc^. 
diameter (that is, the nuniber of times the diameter is 
contained in the circumference) is the same, and may be 
i by an appropriate symbol {ti). 




Proof. 



Given two © with cireumferences denoted by G and C, 
and diametei's by V and D', respectively. 

To prove | -|-» 

C D 

By alternation Jt^T/ Art- 30T. 

Hence, in any given circle, — has a vahic equal to that 



.■. the value of — is the same in all circles. 
Denoting this constant by ti, in every eircle"^ = 7t. 

Q. £. B. 

444. Formula for the circumference in terms of the 
ladius. 



 1) 



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EEGULAU POLYGONS 

445. Formula for length of arc of a circle. 
arc _ central Z 

circiiiufereEoe 360° 



PiiorosiTioN XII. Thkorkh 
446, TJir arra of a regular polyjon is equal to oiie-luilf 
llhe product of its peritiieter hy its iipoihem. 




Given the regnhir polygon ABGJ>E with area denoted 
by K, perimeter by I', and apothem by r. 

To prove £=JPX?-. 

Proof. Draw the radii OA, OB, 00, etc., dividing tho 
polygon into as many A os, the polygon has sides. 

Ail the ^ thus formed have the same altitude, r. Art.308. 

.■. the area of each A = i product of its base by r. 

Art. 3S9. 

Hence the sum of the areas of the & = | product of the 
sum of the bases of the A by r, or= % PX-r. 

But the sum of the areas of the & equals the area of 
the polygon. Ax. 6, 

Hence A'='. V X )■. An. s. 

Q. E, D. 

447. Def. Similar sectors are sectors iu different cir- 
cles which have e{|ual angles at the center. 

448. Def. Similar segments are segments in different 
circles whose area subtend o'liud angles ut the center. 



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TliOrOSITIOX XIII. TUEOKEM 

449. T!ir arm nf a circle is equal to one-half the prod- 
i€t of it-1 circumference hij ils radius. 




Given a O with fiircunifciTiiee denoted Ity C, radius by 
E, and avea h_v K. 

To prove K= i C X B. 

Proof. Cireiimscvibe a regular polygon about the given 
eircle, and denote its perimeter by P and its area by K'. 

In this ease the apothem of the regular polygon is R. 

Hence E'^iPXK. Art. 44fi. 

Let the number of sides of the circumscribed polygon 
be increased indefinitely; then 

iT' becomes a variable approaching £^ as its limit; Art. 441, 

Pbeeomes avariable approaching Gas its limit; Art. 441. 

And viiriiilik'^i'Xfliipproitches I Ox ^iis a limit. Art. 253,2. 
K' = i i' X A' always. Art 446. 



But 
Uenc: 



--\Cx R. 



0. E. I 



450. Formula for the area of a circle in terras of the 
radius li. 

K^i CXB; but C=27lR. Art. 444. 

.-. -i(27r-fi)-K. Or K^TiE''. Ax. e. 

Again E-J i> .-. .fi^^-^-- As. 8. 

461. Cor, 1. The area of a circle is equal to the square 
of the radius, multiplied hy n; or to one-fourth the sgiiare o/ 
the diameter, muUiplied hy %. 



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MEASUEBMENT OF THE CIECLE 277 

452. Cor, 2. The areas of two circles are to each other 

as the squares of their radii, or us the squares of their diam' 

eters. 

K TtIP E- K inJ)'^ D" 

For 



453. The area of a sector is equal to one-half the prod- 
uct of its radius by its arc. 



1 { ceiiti-al I \ 



454. Cor. 3. Similar sedor^ 
squares of their radii. 



Ex. 1. The measurement of tiie area of a circle can be reduced 
to the measurement of the length of what single straight line ! Can 
it he reduced to the measurement of any otiiev single straight line I 
»nt of a sijigle curved line? 



Find the area of a circle, 

Ex. 2, Whose radius is 10 ft. Ex. 6. Whose radius is 2li ft. 

Ex. 3. Whose diameter is 10 ft. Ex. 7. Wliose radius is E; IB; 

Ex, 4. ■\Vhose radius is li ft. Et, 8. Wliose radius is Ei/5. 

Ex. 5. Whose radius is iJ> ft. Ex.9. Whose diameter is *Bi/3. 

Ex. 10. If the radius of oue cirele is 10 times as gfeat as the ra- 
dius of another circle, how do their areas compared Also bjw do 
their circumferouees compare ? 

Ex. II. A wheel with 6 cogs is geared to awheel with 4S eogs, 
How many rovolutious will the sniaUer wlieel make while the larger 
Wheel revolves once ? 

Ex. 12. A 2 in, pipe will diseharKe how much more water in a 
^ven time thau a 1 in, liipij f 



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2 /« BOOK V. PLANE GEOMETliY 

Propositiom XIV. Theorem 

455, 77»' areas of iito similar serjmeutsare tcreach other 
as (he squares of their radii. 



A 



Given the siraiiar segments ABC and A'B'C in ciroici, 
'liose radii are B and R'. 
To prove segment ABC -. segment A'B'C' = B~  It'"^. 
Proof. Draw the raGii OA, OB, O'A', O'B'. 
Then the sectors OAB and (yA'B' are similar. 



And A OAB and O'A'B' are similar. 
. sector OAB R' , A 0.4 R 



I. 448, 447. 
Alii. 327. 



t. 454, 338. 



" scetoc O'A'B' E'-' A O'A'B' 

. seetor OAB _ A OAB 
" sector O'A'B' A O'A'ii'' 
. sector 0-4 B i sector O'A'B' 
" A O^ii A O'A'B' 

. sector OAB— A OAJJ__sector O'A'B'— A O'A'B' 



segment ABO __ segment A'B'C 

A OAB ~ A 0'.d'B' 

;ment ABO _ f A OAB \ ^Kf . 

meat A'B'C \A O'A'B' J W^ 



rts. 307,308. 
Q. £. I). 



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CONSTRUCTION PBOBLBMS 
CONSTRUCTION PROBLEMS 

Proposition XV. Problem 
456. To inscrihe a square in a given circle. 




Let the pupil supply a aolutiou. 

457. Cor. By Msecting the arcs AC, CB, BD, etc., 
and drawing chords, a regular octagon may ie inscribed i» 
the circle; by repeating the process, regular polygons 0/I6, 
82, 64, . , . and 2" sides may be inscribed, where n is a 
positive integer greater than 1. 

How eau a regular polygon of 2" sides be cireuinscribed 
aboiit a given circle 1 



PROPOiilTION XVI. PROBLEJI 

inscribe a regular JK^xagon in a given, circle. 




Let tbe pupil aiipply a Hointion. 



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BOOK V. ri.ASE GEOMETllY 



of a regular innrrihed hexagon 



459. Cor. 1. The si< 
equals the radius of the c 

460. Cor. 2. By joining ihe alternate vertices of a 
regular inscribed hexagon, an equilateral triangle can he in- 
scribed in a given circle. 

461. Cor. 3. By bisecting the arcs AB, BC, CD, etc., 
and drawing chords, a regular polygon of 12 sides can be 
inscribed in a given circle; by repeating the process, regular 
polygons of 24, 48, ... 3 X 2" sides can be inscribed. 

Similarly, how can a regular polygon of 3 X 2" sides 
be circumscribed about a given circle ? 



Proposition XVII. Problem 
462. To inscribe a regular decagon in a given circle. 




Given the circle 0. 

To inscribe a regular decagon in the given O. 

Construction. Draw any radius OA, and divide OA in 
extreme aud mean ratio at K, OK being tbe greater seg- 
ment. Art. 371. 

With A as a center and OK as a radius, describe an arc 
cutting the given circumference at B. 

Then the chord AB J5 a side of the decagon required, 



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CONSTRUCTION PROBLEMS 281 

Proof. Draw OB and KB. 

Then OA -. OE^OK : KA, Constr. 

But AB^OK, .: OA : AB-=AB -. KA. Ax. y. 

Also lOAB^lKAB. idont. 

.-. A OAB aud EAB are similar. Art. 327. 

/. ^0 — ^ABK [homolog. A of similar &). 

And A AKB is isosceles, Art. 321. 

[bdiig simHar lo A AOB, ttfticA is isosceles), 

;. AB^KB=KO. Ax- 1. 

Ileiicu 10= IKBO. (Why?) 

J .-. A0AB=2 10. Art. yti. 

[ 10^ 10. 

Adding, Z AiJO + Z OAS + Z 0- 5 Z 0. 
But Z ABO + Z OAB + Z 0=2 rt Z . Art. 134, 

.-. 5 Z0=2rt. A. Ax. I. 

.-. Z 0=i of 2 rt. A , or iV of 4 rt. A . 
:, are AB is iV of the eiretimf ereneo , 
Hence, if the chord AB be applied ten times in suc- 
cession to the circumference, a regular decagon will be 
inscribed in the given circle. An, 418. 

Q. E. F. 

463. OoR. 1. By joining the alternate vetikes of a 
regular inscribed decagon, a regular pentagon can be inscribed 
in a given circle. 

464. Cor. 2. By bisecting the arcs AB, BC . . . and 
draiving chords, a reyular polygon of .20 sides can he in- 
scribed in a given circle; by repeating iJte process a regular 
polygon of iO, 80, ... 5 X 2" sides can be inscribed. 

How can a regular polygon of 5 X 2" sides be cireuia- 
scribed about a given circle ) 



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28!^ liOOK V. PLASE OEOMKTHV 

Pkoposition XVIII. Pkoblej! 

465. To inscribe a regular polijgon of fifteen sides (pen- 
tedecagon) in a given circle. 




Given the O AK. 

To inscribe a regular pentedecagon in the given O . 

Construction. Draw the chord AC equal to the radios of 
the given O, and the chord AB equal to a side of a regular 
decagon inscribed in the circle. Art. 462. 

Draw the chord BC. 

Then BC is a side of the required pentedeeagon. 

Proof. .AC is a side of a regiiiar inscribed hexagon. 

Art. 459. 
.■, arc AC=i of the circumference. , 
In like manner are AB = iVof the circumference. Art. 4G2. 
.", arc BO—t — "i\, or tV of the circumference. As. 3. 
Hence, if chord BC be applied fifteen times in succes- 
sion to the circumference, a regular pentedeeagon will be 
inscribed in the given circle. Art. 418, 

Q. E. F. 

466. Cor. By bisecting the arcs BO, CD, . . . eic, 
drawing chords, and repeating the process, regular polygons 
of '60, 60, . , . 15 X 2" sides can M inscribed in a given 
circle. 

How can a regular polygon of 15 X 2" sides be circum- 
scribed about a given circle ! 



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COMPUTATION PROBLEMS 283 

COMPUTATION PKOBLEHS 

Proposition XIX. Problem 
467. Given the side and raditis of a regular inscribed 
polygon, to find the side of a regular inscribed polygon ofdouUe 
the nuttier of sides, in terms of the given quantities. 




Given the circle with radius R, AB a side of a regular 
inscribed polygon, and AC a side of the regular inscribed 
polygon of double the number of sides. 

To determine AC in terms of AB and E. 

Solution. Draw the radii OA and OC. 

Then OC is the X bisector of AB. (Why () 

But arc ACB < semieircumferenee. (Why ?) 

.: AC < a quadrant. Ax. 10. 

.*. Zj4 00 is an acute Z. Art. 257. 

Henee,in A 010, Zc^=OP+0C^— 2 OCX 07'. 

Or AC^^2E^ — 2BXOD 

But, in the rt. A OAD, 7)& ^OA^—AlP , Art. 347. 

Or OF = R' — \iABY-. Ax. 8. 

.: OD^V'E-~i JF^i V'ili^ — AB'. 

Substituting for OD its value thus obtained, 



A<f = 2E^ — R-l/i^ —A^. 
AC-= VKi^E—ViB^ — Aff), 



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284 BOOK V. I'LAXE GEOMETRY 

468. Special Formulas. If tho radius B, lie takt>n as 1. 

AC = \^2 — y4: — AB\ 

If a side of tlie regular inscribed polygon of n sides be 

denoted by S„, and ^=1, then S-y,,^^'^ — VT^^Sl 



Proposition XX. Problem 
To compute approximately the numerical value ofn. 




Given a O whose radiiis is 1, and whose eireumferenoe ia 
denoted by C. 

To compute C, i. e., 27t, and henee find the miniorieal 
value of 7t approximatelj-. 

Computation. 1. Inscribe a regular hexagon in the 
given circle, and denote its side by &■ 

Then ySc = l. Art, 459. 

.', perimeter of the inscribed hexagon = 6. 
2. Inscribe a regular polygon of double the number 
(12) sides. 

Then, by the second formula of Art. 4GS, 

Si2=V2- 1/4^^=0.51763809 +. 
Denoting the perimeter of a regular inscribed polygon 
of n sides by P„, 

Pi2 = 12(0.5176380D+) = G.211t!5r08. 



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MAXIMA AND MTKIMA 

S. In the second formula of Art. 468, let h = 12. 



.-. 834=^2—1/4— (0.51763809+)- = 0.26108238+, etc. 
Hence Pw = 6.26525772. 

Computing 8ig, P48, etc., in like manner, the following 
results are obtained; 



5.=l/2- 


-Vi — i 


= .51763P09. 
= .2610523S. 
= .130S0fi2^. 
= .0Gr>43817. 
= .03272346. 
= .01030228. 
= .00S18iaj. 


■- Fit 

: P2 

•■. F, 
.". Fa 

.-. Pi 

-■■ -Pt 
■■• P- 


= 6,21165708. 


Su =1/2- 


-i'4-(.5t76809)' 


= 6.26525722. 


^. =1/2- 


-V4 — {.2til052;(S)' 


= 6,27870041. 


S„ =V'2- 


-V4 — (.13080(i26)' 


= 6.28206390. 


S„.. = VT- 


-V.i~-{.omi3sny 


^ = 6.28290510. 


-s,„=y'2- 


-V4 — (.032733-46 J' 


„ = 6. 28311544. 


sn>=y'2^ 


-V 4 — 1.01636228) 


6 = S.2S31604I, 



Bj' continuing the computation it is found that the first 
six decimal figures in the value of the perimeter of the in- 
scribed polygon remain unchanged. 

.-. C, or 271 = 6,283169 approximately. 
.". 7t=3.14159 approximately, 

MAXIMA AND MINIMA 

470. Def, a maximum (see Art, 268) is the great- 
est of a group of magnitudes, all of which satisfy certain 
given conditions, 

ThuB, the diameter is the maximum oliord wLicli imn be drawn in 

471. Dep. a minimum 13 the smallest of a group of 
magnitudes, all of which satisfy certain given conditions. 

Thus, of all lines which can he dnira from a given point to ft given 
Hue the perpendicular is the miniinum. 

Certain masiiua and minima have already been studied, and we 
now proceed to investigate more particularly those relating to regular 
poljgons and tiie circle. 



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Hi BOOK V. IT AXE GEOMETKY 

Proposition XXI. Theorem 
472. Of all trinngles which have two sides equal, tJutt 
■i'liK/lr ill which these sides include a right angle is the 




Given the A ABC and A'B'C in wliicli AB^A'B', OA 
= C'A', and /.A' is a rt. Z. 

To prove A A'B'C > A ABG. 

Proof. In the A ACB, draw CD 1 AB. 
Then CD < CA. (Why?) 

.-. CD < CA'. Ax. 8. 

But the &. ABC and A'B'C have equal bases. Hyp, 
.■. these ^ are to each other as their altitudes, CD and 
CA'. Art. 391. 

.-. A A'B'C > A ABC. 
(for C'J' > rii). 

Q. E. B. 

473. Iso perimetric figures are figures having equal 
perimeters. 



El. 1. Find the 
iven parallel lines. 



1 between two 



Ex. 2. What is the largest stick that can be placed on a reetangu- 
lar table 12 I 5 ft,, and not have an end projeuting over a side of 
the table f 



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MAXIMA. AND MIKIMA 287 

Proposition XXII. Theorem 
474. Of. all isopfrimetric Mangles which have the same 
ase, the isoscelea triangle is the viaximitm. 




I 



Given the & ABO and AUT) having the same base AB 
and eqnal perimeters, AC=CB, and AD and DB ub- 
equal. 

To prove A ACB > A ADB. 

Proof. Produce AC to S, making CF=AG. 

Draw FB. and from Z) as a center, with a radius equal 
to T>B, describe an are catting FB produced in G. 

DrawDG and AG. 

Draw CS and DK 1 AB; also Gi and DP 1 FG. 

Then Z A£f is a ri?bt Z , Art, 2G1. 

( for it maij be inscribed in a semioirHe whose center is C and 






s ACF). 



Also ADG is not a straight line, 
(/or, if it were, the A DAB and DBA WOiM be complements of the = A 
DGB and DBG, respectwely, and hence icoiikl be equal, aitd .: A DAB 
would be isosceles, which is contrary to the hypothesis). 
:. .1F=AC+ CB=AD-\- DB = AD + DQ. Cocstr. Hyp. 
But AD+ J)G > AG (Art. 02). :. AF > AG. Ax, 8. 
.-. BF > BG. Art, 111. 

.-. i r,F > h BG, or Cfl > KD. Ax. 10. 

.-. A ACB > A ADB. .\yt.:m. 

(J. E, O. 



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288 book v. plane gkomktiiy 

Proposition XSIII. Tueoeem 

475. Of hopfirimetrin pohjgonx haviny the same number 
of Hides, Ihf maximum is eqniUittral. 




Given ABODE the maximum of all polygons having a 
given perimeter, and a given number oE sides. 

To prove -4 BGDB equilateral. 

Proof. If AfiCDE is not equilateral, at least two of its 
sides, as AB and BO, must be uneqnal, 

If this is possible, on the diagonal -iOas a base, con- 
struct a triangle having the same perimeter as ABC, and 
having side AB'^B'C. 

Then A AB'C > A ABC. An. 474. 

To each of these uneqnals add the polygon ACDE. 

:. AB'CDE > ABODE. As. a. 

But this is contrary to the hypothesis that ABODE is 
the maximum of the class of polygons considered. 

Hence AB = BO, and ABODE is equilateral. 

^___ "!- E. ». 

Ei. Of all circles which are described on a given liue aa chord, 



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maxima and minima 289 

Proposition SSIV. Theorem 

4'76. Of all polygons hmnng all sides given hut one, ihe 
■maximum can he inscribed in a semicircle having the unde- 
termined side as a diameter. 




Given the polygon ABCDEF, (he maxitnnra of all poly- 
gons having the sides BC, CD, BE, EF, FA, in common, 
and J.B undetermined. 

To prove that AB is the diameter of a semieircle in 
■which ABCBEF Qs.x\ be inBcribsci. 

Proof. Bravv lines from any vertex, E, to A and B. 

The A BEA must be the maximum of ail A having the 
sides BE and EA, 

(for, if it is nnt, hj incrmsitig or decreasing the angle BEA, thf A KEA 

can be chaitgeA till it is a maximum, ihe rest of figure, BCDE a'nl 

EFA, laeaniime remmniug vnckanged; thus the area of the poly • 

gon ABCDEF iBould 6e increased, wkich is contrary to Ihe 

b'jpotliesis that ABCDEF is a raaxiiHum) . 

.'. Z BEA is a right Z . Art 47^. 

.". E is on the aemicircumference of whiuh AB is the 



In like manner the other vertices, C, D and F, must lie 
on the semicircumference which has AB for a diameter. 



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BOOK V, TLASE GKOMETHY 



Proposition XXV. Thbokkh 



477. Of all polygons formed ; 
that which can he inscribed in a ri, 



iih the same given sides, 
"If in the maxivium. 




Given ABCDEa. polygon wliich cau be iuacribeii iu a O, 
and A'B'CD'E' a polygon wliich has the same sides as 
ABCBE, but which cannot be inscribed iu a O. 

To prove ABGBE > A'B'G'iyB'. 

Proof. From any vertex, A, of ABODE draw the 
diameter AK, and join K to the adjacent vertices C and D, 

Upon CD' constrnet the triangle C'E'D' equal to A CDK, 
and draw A'ff . 

Then area ABGK > area A'B'G'K'. 

Also area AEDK > area A'E'D'K'. 

Adding, ABGKVE > A'B'G'E'D'E'. 

But A CED=A G'K'D'. 

Subtracting, ABODE > A'B'CD'E'. 



Art. ^76. 
(Why t) 
(Why?) 

(Why T) 



Q. E 

478. Note. It might happen that one nf the parts of the soeond 
figure formed by the diameter A'E', aa A'B'C'E', eould be ioEcribed ia 
a Bemicircle, and .". = ABCK. How, then, mowid the above proof be 
modified t 

479. Cor. Of all isoperimftric polygons of a given 
niimier of sides the maximum polygon is regular. 

For it is equilateral (Art. 475), and can be inscribed in 
a sircle (Art. 477), and is, therefore, rfifiiilar (Art, 417). 



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MAXIMA AND MINIMA lifl 

Proposition XXVI. Theorem 

480. Of two isoperimetric regular pnhjrions, that ivhich 
has the greater number of sides has (he greaUr area. 




Given Jta regular polygon of any number of sides, as 
three, and K' a regular polj'gon of one more, or four sides, 
and let K and K' have equal perimeters. 

To prove K' > K. 

Proof, Prom any vertex, C, of K, ili-fiw a line CD to any 
point D of the side AB, which meets one of the sides of 
ZC. 

Construct the A DCF, having CF^DA, and DF^CA. 
:. A DCF= A CDA. Art. I01. 

Adding A CBD, DFCB o K. A^. 2. 

Henee the polygon DFCB has the same perimeter as K' , 
and the same area as K. 

But DFCB is an ii-regukr, while K' is a regular poly- 
gon of four sides. 

.-. K' > DFCB. Art, 470. 

.-. if > A'. Ax. 8. 

In like manner it may be shown that a reguhir polygon 

of one more, or five sides, >K, and so on, 

Q. E. B. 

481. Cor. Of isoperiineiric plane figures, (he circle is 
tile maximum. That is, the area of a circle is greater than 
the area of any polygon with equal peiimeter. 



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292 BOOK V. FLAME GEOJIETEY 

Proposition XXVII. Theorem 

482. Of two equivalent regular polygons that wMch h 
the less number of sides has the greater perimeter. 




e' 




, — p — 

H 



Given ^and E' two regular polygons having the same 
area, and having their perimeters denoted by P and P', but 
£7 having the less number of sides. 

To prove I" > P. 

Proof. Let ^ be a regular polygon having the same 
number of sides as K', and the same perimeter as K. 

Then K > S. Art. 480. 

.-. JU > H. Ax. 8. 

.-. P' > P. Art. 3P9. 



El. 1. Find the area of a trij 
and 12 in., and the im^ludifd Siiig 
triangle having two aides of 6 an 

Ex. 2. How long is the feni 



g!e in which two of the sides are 6 
is 90". Find the area of another 

12 in,, and the inuluded angle 60". 
about a garden 60x40 ft.? How 



many square feet in the area of the garden t Find also the length of 
fence and area of a garden 50 ft. square, 

Ex. S. Find the area of aa equilateral triangle, a square, a heia= 
gon, and a circle, in each of which the perimeter is 1 ft. 

Ex. 4. Find the perimeter ot an eqnilateral triangle, a square, and 
a circle, in each of which the area is 24 sq, in. 

Sx. 6. What principle of ma^sima and minima is illustrated in 
eaob ot the four preceding Exs.? 



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483. Symmetry of polygons. Many of tlie propertiea 
of regular figures can be obtained in a aimple and expedi- 
tious way by the use of the ideas of symmetry. 

484. An axis of symmetry is a line snch that, if part of 
a figure be folded over upon it as an axis, the part folded 
over will coincide with the remaining part of the figure. 



EXERCISES. CROUP 46 

Ex. 1. How many axes of symmetry liaa an 

Ei, 2. How many lias a square? a rcgulai' 
pentagon ! 

Ex. 3. How many lias a regular liesagoii ? a 
rtgutar heptagon f 

Ex. 4. How many haa a regular uctagon ! a 
regular polygon of n sides T 

Ex. 5. How many has a circle f 



485. A center of symmetry for a polygon is a point 
such tliiit jiiiv line di'awn throiigli the point and terminated 
by the perimeter is bisected by the point. 



squa 



EXERCISES. CROUP 4 

equilateral trlanglH a center 

r pentagon a centei' 



;.f syn 



letry f Han a 
iBtry ? Haa a 
enter of sym- 



Ex. 2. Has a n 
regular hexagon ? 

Ex. 3, 111 general, which regular polygons have a c 
metry, and which do not ? 

Ex. 4. Has a circle a center of symmetry t 

Ex, 6. W'liich is the moat eymiuetiical plans figure atudied thus far f 



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EOOK V. TLANE GIlOMliTiiY 




SYMMETKY WITH RESPECT TO A LIME OK AXIS 

486. Two points symmetrical -with respect to a line or 
axis are poiuts such tliat the straight Hue joining ^ijem ig 
bisected by the given liue at right angles. 

Thus, if PP' is bisected by AB, and PP' is 
X AB, the poiuts Pand P' are symmetrical 
with respect to the axis AB. 

487. A figure symmetrical with lespect 
to an axis is a figure such that each point 
in the one part of the figure has a point in 
the other part symmetrical to the given 
point, with respect to an axis. 

488. Two figures symmetrical with respect -^f^^^-^-^^c 
to an axis are two figures such that each point 
in the one figure has a point in the other 
figure symmetrical to the given point, with 
respect to an axis. 

li' 
SYMMETRY WITH EESPECT TO A POITJT OR CEKTER 

489. Two points symmetrical with respect to f^ 
a point or ceater are points such that the / 
straight line joining them is bisected by the / 
point or center. 

Thus, if PP is bisected by the point (7, C is / 
a center of synunetry, with respect to P and P. -y 

490. A figure symmetrical with respect 
to a point or center is a figure such that 
each point in the figure has another point 
in the figure symmetrical to the given 
point with respect to the center. 



49 1 . Two figures symmetrical with respect 
to a center are figures such that each point in 
one figure has a point in the other figure 
symmetrical to it with respect to the center. 



-^'^i?^^ 




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Proposition XXVIII. Theorem 

492. If a figure is symmetrical with respect to tivo axes 
tcMch are perpendicular to each other, it is symmetrical with 
respect to their point of intersectimi as a center. 




Given the figure ABO . . .H symmetrical -with respeot to 
the two axes XX' and YY'; and XX' 1. YY' and intersect- 
ing it at 0. 

To prove ABC ... if symmetrical with respect to 0. 

Proof. Take any point P in the perimeter of the 
figure, and determine the points F' and P", symmetrical with 
respect to P, by drawing FKP' L XX', and PLP" ± YY' . 

Draw KL, P'O. OP". 

Thon Pr [1 i'i", and PP" || XX'. 

But PK^KP. 

Also" Pfi:=and!i LO. 

:. i'P = and || LO. 
:. KLOP is a Cn , andA'i = and || PO. 

In like manner it may be shown that KL = B.ih 
.: PO=nnd II OP". Ax. 



Art. 


121. 


Art 


48fi. 


Art. 


157. 


A 


X. 1. 


Art 


160. 


1 01 




Art 


122. 



Hence any straight line drawn throngh 0, and terminated 
by the perimeter, is bisected at 0, 



. is a center of symmetry for the given figure. Ar!. 4Eio, 



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I'LANE GEOMKTili" 



EXERCISES. CRO 



Ex. 1. The diagonals uf a regular pentagon 
Ex. 2. 1( ABODE is a regular penta' 



.rmnjrit.. 



■lUK a 






Ex.3. Ill the same figure Bt' = irC. 

Ex. 4. Also KCDE is a parallblograiii 

El. 5. Also JC=JB + t!A-. 

Ex. 6. Tlie diagonals of a regular p( 




Ex. 7. The apotliei 
lugle equals one-half  



of a 



Ex. 8. Tho altitude of au equilateral ti 
equals one and a half times the radius of t 
eumserjbed circle. 




Ex. 10. The side of an equilateral triangle equals Ry"i, tl 
oribed circle being denoted by It. 



El. 11. Theapothom 
hiilf the side of a r^uiar 



, regular inscribed hexagon 
rihed triangle { = -x]/'^)- 



Ex. 12. The side of a reg 
side of the regular inscribed triaogle. 

Ex. 13, Find the side of an insc 
the side of a circumscribed square. 

1 of a 

terms of E. 
13 of the ins 



rihed square in tarn 
I inscribed square, i 



ribed and circumscribed squares 
one of these is double the other. 

a of a regular inscribed triangle ia '-'-j — ' 



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ESEKCISES. THEOREMS Ivi 

Ex. 17. Shovf that the area of a regular inscribed hexagon is 
•^ — . What, then, ia the ratio of the area of a regular inaoribed tri- 
angle to that of a regular iusi^ribed hexagon 1 

Ex. 18. The area of a regular inscribed hexagon is three-fourtha 
the area of the regular circumscribed hexagon. 

Ex. 19. The area of a regular inscribed hexagon ia a mean pro- 
portional between the areas of a regular inseribed and a regular cir- 
cumsoribed triangle. 

LSuo. On a figure similar to that of Prop. IV, p. 2(i6. let OP inter- 
seet AB in K, and compare the A OKA, GAP, and OPA' .'[ 

"Ex. 20. The area of a regular inseribed polygon of 2n sides is a 
mean proportional between the areas of regular inscribed and ciruum- 
seribed polygons of n sides. 

Ex. 21. The area of a regular inscribed oetagoa equals the area of 
the rentaugle whose base and altitude are the sides of the oircum 
scribed and inscribed squares respectively. 



Ex. 23. An angle of a regular polygon is the supplement of the 
angle at the center. 

Ex. 24. Diagonals drawn froni a vertex of a regular polygon of n 
sides divide the angle at that vertex into ?i^2 equal parts. 

Ex. 25. The diagonals formed by joining the alternate vertices of 
a regular hexagon form another regular hexagon. Find also the ratio 
of the areas of the two heiagons. 

Ex, 26. If squares be erected on the sides of a regular hessgon, 
the lines joining their exterior vertices form a regular dodecagon. 
Find also the area of this dodecagon in terms of 6, asideof the hexagon, 

Ex. 27. The square of a side of an inscribed e^juilateral triangle 
equals the square of a side of an inscribed square added to the square 
of a side of an inscribed regular hexagon. 



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NR CiEOMETliV 









Ex. SO, Given AaBhC, AcB, BdC, somi- 
eirnles; prove that the sum of the two cres- 
cents AeBa and BdCb equals the ateaoE the 
right triangle ABC, 

Ex. 31. An equiangular poly)?oii in- 
scribed ia a oircla is regular if the number o£ it 
[SUQ. In the figure to Prop, III, p. 264, t 
:. are AE^&sc BO, .". aide Jif^side BC, ete,] 



sides be odd. 

^ JEUC = ai-e EDOB 



EXERCISES. GROUP M 

JlASl.MA A^^D MINIMA 



Ex. 2. Of equivalent piirallelogra 



Ex, 3, Of i 30 peri metric ri 
Ex. 4, Divide a giveu lir 



■s whicli 13 t 



Ex, 5. Find a point in the hypotenuse oi 
the 8um of the squares of the perpendieular 
the legs shall he a minimum. 

Ek. 6. How shall a mile of wire fence 



t triangle such thnt 
3 from the point to 



Ex. 7- Find the area in acres included by a mile of wire (enae if 
it be stretched as a square, a regular hexagon, and a circle respectively. 

Ex, 8, Of all triangles with the same base and equal altitudes, the 
isosoelcs triangle has the least perimeter, 

Ex. 9, Of all polygons of a given number ot aidta inscribed in a 
given circle, the maximum is regular. 

[SuLi. Prove the maximum polygon (1) equilateral, (3) equi- 
angulaj,] 



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EXEIJCTSES. SVMMETKY 299 

E3. 10. Find tlie masimiuii leetatigle inscribea in a circle. 

Ex. 11. Find the maximum rectaugle that eaii bo inscribed iu a 

[Sue. Inscribe a square in tbe circle.] 

Ex. 12. Of trapsKoida inscribed in a semicircle (liaviug the diam- 
ter us one base), find the maximum. 

Ex. 13. Divide a given straight line into two pnrts such that the 
am of tiie aquarea ol these parts sliall be a minimum. 



EXERCISES. CROUP SO 

SYMMETRY 

Ex. 1. A rhombus has how many axes oi! symmetry? Has it a 
center of symmetry ? 

Ex. 2. "What asls o£ symmetry has a quadrilateral which has two 
pairs of equal adjacent sides f Has such a figure a center of symmetry ? 

A parallelogram is symmetrical with respect to Ihe point 
.1 — of its diagonals. 



Ex. 4. A segment of a circle is symmetrical with respect to wha 
xis! 
Ex. 5. Has a trapezium a center of symmetry ! An axis o 



Ex. 6. How many axes of symmetry have two equal circles take 
as one figure 1 Have they a center of symmetry I 

Ex. 7. "What axis of symmetry have any t*o circles f 

Ex. 8. How must two equilateral triangles be placed so as to hav 
& center of symmetry f So as to have an axis of symmetry f 

Ex. 9. It two polygons are symmetrical with reference lo a een 
ter, any two homologous sides are equal and [>arallel and drawn i 
opposite direetioua. 



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lOK V. I'LANE GEOMETUi 

EXERCISES. GROUP 81 

VARIOUS THEOREMS 

if a eirtle equals 'wice tlie 

an inscribod equilateral triangle la to the 

Ex. 3. Tho diagoaal joining any two opposite voL'litts of a regulaf 

Ex. 4. Tha radius of an inaeribed regulsL' polygon is a mean pro- 
portional between its apotbem and the radius of tlie uireumacribad 

regular polygon of the same number of sides. 

Ex. 5. If the sides o£ a regular hexagon be produced, their points 
of intersection are the Tortices of another regular hexagon. Also Had 
the ratio of the areas of the two hexagons. 

Ex. 6. Of all lines drawn through a given point within an angle, 
and terminated by the sides of the angle, the line which is bisected at 
the given point cuts off the minimum area. 

Ex. 7. Eaeh angle of a regular polygon of n + 2 sides contains 
_ rjgiit angles. 

Ex. 8. The diagonals from a vertex of a regular polygon of ii -[- 3 
sides divide the angle at that vertex into « equal parts. 

Ex. 9. The sum of the perpendiculars drawn to the sides of a 
regular polygon of n sides from any point within tho polygon e(|uals n 
times the apothem. 

Ei. 10. An equiangular polygon circumscribed about a circle is 

[SUQ. Draw radii from the points of contact, and lines from the 
vertices of the polygon to the center.] 

Ex. 11. An equilateral polygon circumscribed about a polygon is 
regular if the number of its sides is odd. 

[Sua. See Figure of Prop- III, p. 264. Prove Jr=iiy, Draw radii 
and prove /.T= IQ, etc.] 



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EXEECISES. MISCELLANEOUS THEOREMS 301 

Ex. 12, How must two equal isosceles triangles be placed so aa 
to have a center of symmetry f How masl they be placed so as to 
have an axiS of symmotry ! 

If, in a regular inscribed polygon of n sides, •%, denotes a side and r™ 
denotes the apothem, show thjil. 

Ex. 13. For the triangle, S3 = Rv''3, r,, = i5. 

Es. 14. For the square, S, = Bv'2, i-. = iJtv'3. 

Ex. 15. For the heiagoa, S6 = K, r6 = iBv'3- 



Ei. 18. For the octagon, Ss~RV'2—i/2. 

Ex. 19. For the dodecagon, Su =• EV'2—t/Z. 

Ex.30. Prove that V = K' + Sio'. 

Ex.21. Trove that Si,' = S„' + S|f,'. 

Ex. 22. IE ADB, AaC, CbB are semicir- 
cles and DC ± AB, prove that the area 
bounded by the three semicircumferencea 
equals the area deaoribed on I>C as a diameter. ■^ ij Js 

Ex. 23. If p„ denotes the perimeter of an iaseribed polygon of n 
sides and P„ the perimeter of a oitonmaoribed polygon of n sideB, 
1p^ P^ 



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302 BOOK V. I'LAKE GEOJIETIii 

EXERCISES. CnOUP S2 

PROBLEMS 
Circumapribe about a ftivon eire5e 
Ex. 1, An equilateral triangle. 
Ex. 2. A square. 
Ex. 3. A regular p':^ntagon. 
Ex. 4. A regular hesagou. 

Ek. 6. A regular dpcagon. 

Ex. 7. Constpuet a regular pentagram, or .i-poinled star, 

Ex. 8. Conatruet a he.Yagram, or O-pointed star. 

Ex. 9. Constniet an S-pointed star. 

Ex. 10. Construct iin angle of 36". 

Ex. 11. Construct angles of 18', 9", 72°. 

Ex. 12. Construct angles o£ '2i'', 12°, G", 48°. 

eireumferenpe into two parts which shall 

Ex. 14. Construct a regular pentagon which shall have twice the 
perimeter of a given regular pentagon. 

Ex. 16. Conatruet a regular pentagon whose perimeter shall equal 
the sum of the parlmeters oC two given regular pentagons. 

Ex. 16. CocBtruot a regular pentagon whose area sliail be twice 
the area of a given regular pentagon, 

Ex. 17. Construct a regular pentagon whose area shall be equal 
to the sum of the areas of two given regular pentagons. 

Ex. 18. Constru 
oamferenee of a giv 



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EXERCISES. PROBLEMS 303 

Ex. 19. Construct a circle whose area shall be three timeH the 
area of a given circle. 

Ex. 20. Construpt a ciroumterenee equivalent ti the =iim and 
another equivalout to the diBereneo, of two gi\en i irtmnleiencoa 



] five equal parts by 



Ex. 


21. 


Construe 


a circle e 


equivalent to th 


e difference, o£ tv 


Ex. 


22. 


Construe 


a circle wh 


ofagi 


e„ 


irole 






Ex. 


23. 


Bis 


ct t!i 


B iiroa of a 


ference. 








Ex. 


24. 


Divide th 


e area of a 


drawin 


geo 


ncen 


rice 


rcnmferene 


Onag 


ven 


line 


cons 


ruct 


Ex 


25. 


Ar 


gular pentagon. 


Ex 


26. 


Ar 


egula 


hexagon. 


Ex 


27. 


Ai 


egula 


dodeeagon 


Ex 


28. 


Ac 


rcle 


qiiivaleiit t 


Ex 


29 


Ins 


criLe 


a regular oo 


Ex 


30 


lus 


cribe 


a circle in a 


Ex 


31 


Ins 


eribe 


a square in 



mieircle. 
;iven square. 



Ex. 32. In a given equilateral triangle inscribe throe equal circlsa 
each of which touches the other two circles and a side of the triangle 

Ex. 33. In a given circle inscribe three equal circles which shal 
touoh eaeb other and the given ci renin ferenoe. 



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NUMERICAL APPTJCATTOXf! OF PLANE 
OEOMETEY 

METHODS OF KUMEEICAL COMPUTATIONS 

493. Cancellation. In numerical work in geometry, aa 
elsewhere, the labor of computations may frefjuentlj- be 
economized. Those methods of abbreviating work, which 
are particularly applicable in the ordinary numerical appli- 
cations of geometry, may be briefly indicated, as follows: 

To simplify numerical work by caneeliation, group 
together as a whole all the numerical processes of a given 
proilem, and make all possible cancellations before proceed- 
ing to a final nuvterical redtiction. 

Ex. YinA the ratio of tlie area of a rectangle, whose base and alti- 
tude are 42 and 24 inches, to tlie area of a trapezoid, whoae bases are 
21 and 35 and altitude 12. 

Bj Arts. 383, 394 J 

area of i-eetanRle ^ >2X 24 _ ^ ^fk?* ^r, j^,. 
area of trapezoid 6(214-35) ^X^B 



494. Use of radicals and of 71. Where radicals enter in 
the course of the solution of a numerical problem, it fre- 
quently saves labor not to extract the root of the radical till 
the final answer is to be obtained. 

Ei. 1. Find the area of a circle oiroumacribed about a squara 
whose side is 8. 

The diagonal of the square must be 8i/2 (Art. 34C). 

.-. the radius of O =4i/2. 

.'. by Art. 449, area of G = x(4i/2)' = 32ir = 100.6. Area. 



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NUMERICAL COMPUTATIONS 305 

Similarly in the use of tt, it frequently saves labor not 
to substitute its numerical value for n tilt late in the process 
of solution. 

Ei. 2. Fiud the radius of a circle whose acefi is equal to the suia 
of the areas of two circles whose radii are 6 and 8 inches, respectively, 

Deaote the radius of the required eirele by x. 

Then, by Art, 440, ^ j' = 36 tt + C4 ,r. 

.-. Tl'= 100 TT. 

,-. j' = 100, and x= 10, Fadiu.^. 

495. Use of x, j, etc., as symbols for unknown quan- 
tities. In some cases a numerical computation is greatly 
facilifated by the use of a specific sijwhol for an imknoirti 
quantity. 

Ex. In a triangle whose aldea are 12, IS, and 2.'), find the segmeots 
of the side 25 made by the bisector of the angle opposite. 
Denote the required segments of Bide ^5 by 

Then 12 : 18 = j: : 2;') — j:(Art. 332) 

.-. 18j- = 12(2.j-j.-) (Art. 302) 
.■.:r = 10.i 
And 25~a: = lf..)" ' 



r Sc(jmfiits . 



496. Limitations of numerical computations. Owing to 
the limitations of human eyesight and of the instruments 
used in making measurements, no measurement can be 
accurate beyond the fifth or sixth figure; and in ordinary 
work, stteh as is done by a carpenter, measurempnts are 
not accurate beyond the third figure. As all numerical 
applications of geometry are based on practical measure- 
ments, it is not necessary to carry arithmetical work beyond 
the fifth or sixth significant digit. 

Other methods of facilitating numerical computations, 
as by the use of logarithms, are beyond the scope of this 
book. 



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6Ub PLANE GEOMEIKY 

KUMEBICAL PKOPEKTIES OF LINES 
EXERCISES. CflOUP S3 

THK KlfiHT THIANGLU 

Ek. 1. Find Uie hypotenuse of a riglit lria»i;lo wht.so leE3 ave 13 
and :i5. 

Eji. 2. The hypotenuse of a right triangle i? 29. and one leg ia 
20. Find the other leg, 

Ei. 3. If a window ia 15 ft. from the (■round and the foot of a 
ladder is to be 8 feet from the house, how long a. ladder is necessary 
to reaeh the window P 

Ex. 4. Find the diagonals of a rectangle whose sides are .j and 13. 

Ex. 6. Find the diaconnl of a square whose side is I ft, 6 in, 

Ex. 7. The diagonals of a rhombus .ire '2i and 10. Find a side, 

Ex. 8. One side of a rhombua is 17 and one diagonal is 30. Find 
tiie other diagonal. 

Ex. 9. In a circle whose radiua is 5, find the length of the longest 
and shortest chords through a point at a distance 3 from the center. 

Ex. 10, In a circle whose radius is 25 in,, find the distance from 
the center to a chord 48 in. long. 

i in. fio 

Ex. 12. A ladder 40 ft. long reaches a window 20 ft. high on one 
Bide of a street and, if turned on its foot, reaches a window 30 It, high 
on the other side. How wide ia tlie street J 

Ex. 13. If one leg of a right triangle ia 10 aud the hypotenuse is 
twit;e the other leg, hnd the hypotenuse. 

Ex. 14. Find the altiludo of an equiiaterai triangle whose side isG, 

Ex. 15, Find the aide of an equilateraltriaugle whose altitude is 3, 



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NUMESICAL EXERCISER. LINES .307 

Ex. 16. Find the side of & square whose diagonal ia 15. 

Es. 17. One leg of a right triangle is 3, and the sum of tha 
hypotenuBs and the other leg is 9. Find the sides. 

Ex, 18. A tree SO ft. high ia broken off 40 ft. from the ground. 
How tar from the foot o£ the tree will the top strilce V 

Ex. 19, The radii of two oirolea are I and 6 in., and their oenters 
are 13 in. apart. Find the length of the common esternal tangent. 

Ex. 20. TiiG Bides of a triangle are 10, 11, 12. Find the length of 
the projection of the side whose length is 10, on tlie side 12. 



EXERCISES. CROUP S4 

TRIANOJ-ES IN GENERAL 

Ex, 1. The sides of a triangle are 13, 18 and 20. Find the aeg- 
fflents of the side 20, made by the bisBOtor of the angle opposite. 

Es. 2. In the same triangle, find the segments of the Bide 20, made 
by the bisector of the exterior angle opposite. 

Ex. S. If the legs of a right triangle are C and 8, find the hypote- 
nuse, the altitude on the hypotenuse, and the projections of the legs 
on the hypotenuse. 

Ex. 4. Is a triangle acute, obtuse, or right, i£ the three sides are 
5. 12, U; 5, 11, 12; 5, 12, 13; 4, 5, 6? 

i are G, T and 8, compute the length 

Et. 6. Also the length of the median on the same side. 

Ex. 7. Also thu length of tlie bisector oE the angle opposite the 
iiJe B. 

Ex 8. It two Bides and a diagonal of a paiulloiotsram ftre 8, 13 
^ia 10, find the other diagonal. 
[SuG, Uao Art. 352.] 



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PLAKE GEOMETRY 



El. 11, The hypoteniisB of a right triangle 'in 10, and the altitude 
on the hjpotenuaa is 4. Find tho aegments of the hypotenuse and tlie 
legs. 

El. 12. Find the three medians, the fliree bisectora, and the tLreu 
altitudes ol a triangle whose sides are 13, 14, 15, 



EXERCISES. CROUP SS 

CIRCTTMFRRENCF.S AM> ARCS 

T.  - 55 
Lamg fl-^.-yt 

Ex. i. Find thn ciroumEerence ot a eirclf whose radius in I ft. 9 in. 

Ex 2. Find the radiuB of a circle whose Rireumferenee is 121 ft. 

Ex. 3. A biojele wheel 28 in. in diameter makes, in an afternoon, 
3,000 revolutions. How far does the bicycle travel F 

Ex. 4. What is the diameter of a wheel whkh makes 1,400 revolu- 
tions in going B,800 yds. ? 

Ez. 5. If the diameter of a circle is 20, find the length of an arc of 
60°; also of 83°. 

Ex. 6. If the length of an arc is 14 and the radiua is G, End the 
namber of degrees in the arc. 

Ex. 7. If the arc of a quadraut m 1 ft. in length, find the diameter. 

Bk. 8, Two eonoentrie circumferences are 88 and 132 in. in length, 
respectively. Find the width of the Bircular ring between them. 

Ex. 9. If the year be taken aa 365i da., and the earth's orbit a 
circle whose radiua is 93,250,000 miles, find the velocity of the earth 
in its orbit per second. 



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NUMERICAL EXERCISES. LINES 309 

Find the r&dius and circumference of a circle circumscribed about 
Ex, 10. A square whose aide is 5. 
Ex. 11. An equilateral triangle nhose side is 4. 
Ex. 12. A rectangle whose sides are 12 and 5, 



Ex. 14. Find the diameter of a cirt 


;le eircH 


angle whose sides aro 7, 15, and 20, 




Ex. 15. Find the radius of a oirclf 


i whose 


the perimeter ol a square whose diagoni 


il is lU. 



EXERCISES. CROUP liS 

CBOKDS, TANGENTS, AND SECANTS 

ES. 1. Two intersecting chords of a cirule are 11 and 14 in., and 
the segments of the first chord are 8 and 3 in. Find the segments of 
the second chord. 

[St;G. Denote the required segments bf :c and 14 — e,] 

Ex. 2. In a circle whose radius is 12 in., a chord 16 In. long is 
passed through a point 9 in, from the center. Find the segtnents of 
the chord. 

Ex. S. Two secants drawn from a point to a circle are 24 and 37 
in. long. If the external segment of the first is G in., find the external 
segment of the second. 

Ex. 4. From a given point a seeant whoso external and internal 
segments are 9 and 16 is drawn to a circle. Find the length of the 
tangent drawn from the same point to the circle. 

Ex. 5. Frenj a given point a tangent 24 in. long is drawn to a 
circle whose radius ia 18 in. Find the distance of the point from the 
center. 

Ex. 6. I( a diameter GO in. long is divided into 5 equal parts by 
«l»ords perpendicular to it, find the lungth of tiu) chowlB, 



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PLANE GEOMETKY 



Ex. 8. If the earlh is 
will tbe light of a HgLlhou 



EXERCISES. CnOtIP Bt 

LINES IN' SLMILAli FICURiiS 

Ex. 1. K the Hides of a triiwgle are fi, 7 and S, and the shortest 
Bide of a similar triangle is 18, find the other aidea of tbe secoLd 

Ex. 2. If a post 5 ft. high easts a shadow :! ft. lotij;, find the 
height of a ateeple wkiph easts a shadow 90 I't. long. 

Ex. 3. In a triangle whose base is 14 and aliitude 1!!, a line is 
drawD parallel to the base and at a distauee 3 froni the base. Find 
tbe length of tbe line thus drawn. 

Ex. 4. The upper and lower bases of a trapezoid are 13 and % 
and tbe altitude ia H. If the legs are produced till they meet, find 
the altitnde of eaph of tbe two trianglts thus formed. 

El. 5. If the upper .-ind lower bases of a trapezoid are i, and li, 
and the altitude ia h, find the altitude of each of tbe triangles formed 
by producing the legs. 



Ex. 7. If the perimeter of a legular poljgon is three times the 

the ratio of their apoihema t 

Es. 8. It the eireumterenees of two eirelea are 600 and 400 ft., 
wbat is the ratio of their diameters 1 

Ex, 9. In the preceding example, if a ohord of the first circle ia 
30, what is the length of a chord in the second circle, subtending 
the same number of degrees of arc 1 



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NUMERICAL EXEECISES, AREAS 6 

COMPUTATIOW OF AREAS 
EXERCISES. CROUP iiS 

AREAS OF TRIANGLES 

i of a triangular field whose b 

El. 2. Find the area oC a triangle wlioae aldea are 10, 17, and 

El. 3. Find the area in aerts ot a liuld whose sides are (iO, 
ud no chains. 

d of a triaag-ulur tield t^xch i>S «li 
clialcs. 
Find the area of 

Ex. 5. Ad isosaelea triangle whose base \3 16, and eai^li of wh 
legs la 34. 

Ex. 6. Aa equilateral triangle whoEe altitude is 8. 

Ei. 7. A right triangle in which the segments of tha hjpotec 
made by tlie altitude upon it are 12 and 3; also, in one in which 



Ex. 8. An isosceles right triangle whose hj-potennse is 12. 

Ex. 9. A right triangle in which fhe hypotenuse is 41 and one leg 
is 9. 

Ex. 10. Find in two ways ibe area of a triangle whose sides are 
6, 5, 5. 

Ex. II. A side of a given equilateral triangle is 4 ft. longer than 
the altitude. Find the area of the triangle. 

Ex. 12. The area of an isosceles triangle is 144 and a leg is 24. 
Find the base. 

Ex. 13. The area of an equilateral triangle is 4t/3. Find a side. 
Ex. 14. The area ot a triani;ie is 1125, and if.h: f = 2 : ;i : 4. Find 

Ex. 15. The area o£ a triangle is G sq. in., and two ot its aides are 
il aud 5 in. Find the remaining eida. 



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i'LANE GEOMETHY 



EXERCISES. GROUP SB 




AREAS OP OTHER RECTILINEAIE t 


■ir.URES 


PimJ tlie area of 




Ex. 1. A par«.lleloi;ram whosa Ijaso is 2i ft. fi ' 
tuaei8l2ft. Oiu. 


ill. and w 


Ex. 2. A trapt^zoia whose uasvs are 12 ami 20 
tude is n ft. 


in. and w 



Ex. 3. A vhombus whoae diagonals are 9 ft. and 2 yds. 

Ex. 4. A quaJrilateral in wliieh the sides All, BC, CD, DA a 
13, 14, 15 and the diagoual AC is 17. 

Ei. 5. A quadrilateral iu which the sides ai-o £7, 36, 30, 2 
the angle included betweeu the first two siciea is a right angle. 

Eit. 6. A square whose diagonal is 12 in. 

Ex. 7. Piud the number of boards, ea<;li 4 yds. long and 
wide, which are neeessarj to cover a floor 48 X 24 ft. 



Ex. 10. A rectangular garden contains 4,524 aq. yds. and 13 20 
yds. longer than wide. Find its dimensions. 

Ex. 1 1. Each Bide of a rhombus is 24 ft. and each of the larger 
angles 13 double a smaller one. Find the area. 

Ex. 12. Find the area of a rhombus one of whose sides is 17, and 
one oC whosB diagonalB Is 3U. 

Ex. 13, The area of a trapeaoid is 4 acres, one base Is J2U yds., 
and the altitude is 100 yds. Find the other base. 

Ex. 14. The bases of an iaoseeles trapei^oid are 20 and 30 and the 
legs an IT. Find the area. 



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NUMERICAL EXEKCISES. AREAS 313 

Ex. 15. The base of a triangle !a 20 and tbe altitude 13. Find 
the length of a line parallel to the base ivhieh cute off a trapezoid 
vi\ioRe area is 80 sq. £t, 

[Suo, Denote the altitude ot trapezoid by 18— a; and Gad its 
•■ppBr base by similar triangles.] 

Ex, 16. The perimeter of a polygon, circumscribed about a circle 
trhose radius is 20, is 340, Find the area of the polygon. 

Ex. 17. The area of a rectangle is 144 and the base is three times 
the altitude. Find the dimensions. 

Ex. 18- Find theareaof aregularhesagon oneof whose sidesialO. 

Ex. 19. Find the area of a regular deeagou inscribed in a eircie 
whose radius is '20. 

Ex. 20. f iud a aide of a. regular hesagoii whose area is 200 sq, in. 



EXERCISES. CROU? 60 

AREAS OF CIRCULAR FIGURES 
Ex. 1. Find the area in aeres of a circle whose radius is 100 yds, 
Ex. 2. Find the radius in inches o! a circle whose area is 1 sq. yd. 
Ex. 3. Find the area ot a circle whose circumference is j). 



Ex. 5. Find the radius of a circle whose area equals 
^e areas of three circles whose radii are 20, 2H, 10. 

Ex. 6. In a eircie of radius 50 find tbe area of a sector 



Ex. 8, IQ a circle whose radius is 7, the area of a sector is 43 sq. ft. 
Find the number of degrees in its angle. 

Kx. 9. Ill a circle whoso radius is 10, fiud the sum ot the segmeiitB 
formed by au inseribed square. 



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PLANE GEOMETHV 



Ex. li. The same pond is surrounded by a driveway 30 ft. wide. 
Find the area of the driveway. 

Ex. 12. Two tangents to a nirele, whose radius is 15, include an 
angle of 60°. Find the waa Included between the tangents and the 
radii to the points of contact. 

Ex. 13. Find the length of the tether by which a cow must be 
tied, in order to graze over eiiactly one acre. 

Ex, 14, Three equal eii'cles touuh each other externally. Show 

that the area included between them iaJtHyS — ^)- 

Kx, 15, If the area inclnded between three equal circles which 
touch each other externally is a square toot, find tlie radius of each 



EXERCISES. CROUP ei 

AREAS OF SIMll.AK FIGURES 

Ex. 1. The homoloKons sides of two similar triangles are 3 and 5. 
Find the ratio of their areas. 

Ex, 2. The liomologoua sides of two similar polygons are 4 and 7, 
and the area of the first polygon is 112, Find tlie area of the second 
polygon, 

Ex.3. The radius of a circle is e. Find rlie radins ot a circle hav- 
ing three times the area of the given circle. 

Ex, 4. The areas of two circles are as IC to p, and the radius of the 
first is 8. Find the radiua of the second. 

Ex. 6. The sides of a triangle are 5, G, 7. Find the aides of a 
similar triangle containing 9 times the area of the given triangle. 

Ex. 6. If, ia finding the area of a circle, a student uaeB 1> = 50 as 
K = 50, how will the area as computed differ from the correct area 1 



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MISCELLANEOUS NUMERICAL EXERCISES 

Et. 7. In a triangte whose base \a 24 in. and altitude is 18 ii 
ftltituiie is bisected by a line purallel to the base. Find tbe a 
the triangle cut off. 



Ex. 9, In a circle wboBe diameter is 30 in,, what are the diameters 
of ooncentric eireumferences which divide the area into three equiva- 
lent parts T 

Ex. 10. If a circle be eoostrueted on the ridius o! a given circle, 
Hud segments, one in each eirGle, be formed by a line drawn from the 
point ol contact, find the ratio of the , 



EXERCISES. CROUP 63 

OENERAL NUMERICAL EXERCISES EH PLANE OEOMETRY 
iiid the base is IQ. 

e equivalent to this triangle. 



Ex. 4. The Rides of a triangle arc 7, 8 and S inches. Find the 
aides of a triangle of four times the area. Also, of twice the area. 

Ex. 5, The temple o! Herod is said to have aooommodated 210,000 
people at one time. If each person required 27 X IS in., and one-third 
the space iiiaide the temple be allowed for walls, sanctuaries, etc., 
what were the dimensions of the temple, if it was a square f 

mgle are 12, 16 and 21., vrhat are the 
/ the bisector of the angle opposite t 

lateral in order 
m aiif;le of (10°, 

Find the diameter o£ a wheel which, in a mile, maires 480 



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PLASE GEOMETKV 



Find the radius of a circle equivalent to 
Ex. 10. A square mlioae side is 10. 
Ei. II, An eijiiilateral triangle whose siile Is 12. 
Ex. 12. A trapezoid whose bases aro IG and 18 and altitude 9. 
Ex. 13. A aemicirelo whoso raijins is 15, 

'!■ of ft circle whose area shall be e<]uiTa- 
I whose diameters are U4 and 17. 

Ek. 15. A oirele, a square, and an equilateral triangle each, have 
s perimeter of 12 yds. Find the area of eai^h figurii. 

Ex. 16. In a circle whose area is 400, the area of a seetor is 125. 
Find the angle of the sector. 



Ex. 17. How many acres are included w 
track, if the traek is in the shape of a rei 
it IE wide 1 



Ex. 19. Find the area of the circle circumscribed about a 
angle whose sides are 40 and 9, 



Ex. 20. One leg of a right triangle 
teen the hypotenuse and the other leg 

Ex. 21. Find the area of an isosceles 



Ex. 22. In a triangle whose sides are 16, 18, 20, find the length of 
the altitude, median, and bisector of the angle opposite the longest 

Ex. 23. A line 16 inches long is divided internally in the ratio of 
3:5; find the segments. Also find the segments when the line is 
divided externally in the same ratio. 

Ex. 24. A line 16 Inches long is divided in extreme and mea& 
ratio. Find the segments. 



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MISCELLANEOUS NOMEEICAI, EXERCISES 317 

Ex, 25. If a line is divided in extreme and mean ratio and the 
Bmaller segment ia 4, flnd tlio whole line. 



Ex. 28. In a circle whose diameter is 20, a chord is passed throuKh 
a point at a distance 6 from the center, perpendicular to the diameter 
through that point. Find the length ot this chord, and ot the chords 
drawn from its extremities to the ends of the diameter. 

En. 29. Eiioh Itg of an isosceles trapezoid is Jll, and onH base 
exceeds the other by 16. Find the altitude. 

Ex, 30. It three arcs, each of 60° and having 10 for a radius, are 
to the other two, find the area included by them. 



Ex. 31. Find the area of a trap? 


izoid whose legs i 


ire 4 and 5, and 


whose bases are 8 and 11. 






Ex. 32. A square piece o£ land : 


and a<-irpnlar pi, 


ice each contain 


1 acre, llow many more feet of fenc 


does one require 


. than the other ? 



El. 33. IE the base of a triangle is doubled and the altitude re- 
mains unchanged, how is the area affected ! If the altitude is doubled 
and the base remains unchanged t It both the base and the altitude 
are doubled f 

ot 



Ex. 37. Find the side of an equilateral triangle equivalent to a 
oitcle whose diameter is 10. 

Ex. 38. The area of a rhombus ia loti sq. in., and one side is 1 ft. 
1 in. Find the diagnnuls. 



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318 PLAN'E GEOMETRY 

Ex. 39. Tlie EJiipB of atriansle are S, in, 12, Find the aicis of 
the triangles iii.icle by llie bisector of the angle opposite Ihu side 12. 

Ex. 40. In a circle of area 275 sq. ft., a re<^tangie of area 150 aq. ft. 
is inaeribed. Sliow how to find the sides of the reetaogle. 



EXERCI3E3. CROUP 63 

EXERCISKS INVOLVING . THE METRIC SYSTEM 
Ex. 1, Find the area of a triangle of which the baee is 16 dm. and 
the altitude SO cui. 

Ex. 2. Find the area of a triangle whofe sides are 6 m., TO dm., 
800 cm. 

Ex. 3. Find the area in square meters of a circle whose radius is 
14 dm, 

Ex. 4. If the hypotenuse of a right triangle is 17 dm, and one log 
iB 150 cm., find the other leg and the area. 

Ex. 5. It the circumference of a circle \i 1 m,, find tlio area of 
the circle iji square deeimeters. 

Ex. 6. Find the area in heetares, and also in acres, of a circle 
whose radius is 100 m, 

Ex. 7, If the diagonal of a rectangle is 35 dm, and one side is 800 
mm., find the area in square meters, and also in square inches. 

Ex, 8, Find the area of a trapezoid wiiose bases are 600 em. and 
2 m., and whose altitude is SO dm. 



Ex, 10. In a given circle two chorda, whose lengths are 15 dm. 
and 13 dm,, intersect. 3f the segments of the first ehord are 12 dm. 
and 3 dm., find the segments of the second chord, 

Ex. 11. Find in det imetera the radius of a circie equivalent to a 
square wbosa side is 1 ft. 6 in. 

Ex. 12. Find in feet the diameter of a wheel which, in going 10 
kilometers, makes 5,000 revolutions. 



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SOLID GEOMETRY 

Book VI 

LINES, PLANES AND ANGLES IN SPACE 

DEFimTIONS AND FIRST PRINCIPLES 

497. Solid Geometry treats of the properties of space of 
three dimensions. 

Many of the properties of space of three dimensiona are determined 
by use of the plane and of the pcopertiea of plaue figures already 
obtained in Plane Geometry. 

498. A plane is a surface such that, if any two points 
in it be joined by a straight line, the line lies wholly in the 
surface. 

499. A plane is determined by given points or lines, if 
no other plane can pass through the given points or lines 
"without coinciding with the given plane. 

500. Fundamental property of a plane in apace. A 
plane is determined by any three points not in a straight line. 

For, if through a line con- ,c 

necting two given poiuts, A 
and B, a plane be passed, the 
plane, if rotated, can pass 
through a third given point, 
C, in but one position. ■'* 

The importance of the above principle is seen from the faet that it 

reduces an unlimited surface to three points, thus making a vast 

econoijij to the attention. It also enables us to eouuect different 

planes, and treat of their properties aystematically. 

(3iy) 



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320 BOOK VI. SOLID GEOMETRY 

501 . Other modes of determining a plane. A plane may 
also be delerm'uied by any equivalent of three points not iu 
a straight line, as by 

a straight line and a point outside the, line; or hy 

two intersecting straight lines; or by 

two parallel straight lines. 

It is often more eonvenient to use one o£ these latter raethods of 
determining a plane than to reduea the data to tliree points md use 




502. Representation of a plane in geometric figures. In 
reasoning concevuiug the plane, it is often an advantage 
to have the plane represented in all directions. Hence, in 
drawing a geometric fignre, a plane is usually represented 
to the eye by a small parallelogram. 

This ia virtually a donble aae of two intersecting lines, or of two 
parallel lines, to determine a plane (Art. 501]. 

503. Postulate of Solid Geometry. The principle of 
Art. 499 may also be stated as a postulate, thus: 

Through any three points not in a straight line ior their 
eqnivaJeni) a plane may be passed. 

504. The foot of a line is the point in which the line 
intersects a given plane. 

505. A straight line perpendicular to a plane is a line 
perpendicular to every line in the plane drawn through its 
foot. 

A straight line perpendicular to a plane is sometimes called a normal 



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LINES AND PLANES 321 

506. A parallel straight line aud plane are a Hue aad 
plane which cannot meet, however far they he produced. 

507. Parallel planes are planes wliich caonot meet, 
however far they he produced, 

508. Properties of planes inferred immediately. 

1. A atraighl line, not in u yircti plane, can in!erseci IJie 
given plane in but one point. 

For, if the line intersect the given plane in two or more 
points, by definition of a plane, the line must lie in the 
plane. Art. 498. 

2. The intersection of two planes is a straight line. 
For, if two points common to the two planes be joined 

by a straight line, this line lies in each plane (Art. 498) ; 
and no other point can be common to the two planes, for, 
throagh a straight line and a point outside of it only one 
plane can be passed. Art. soi. 



Ex, 1, Give an esample o£ a plane Burfa<ie; of a eiu'vod siirfaee ; 
of a Burfaee, part plane and part curved; of a surface composed of 
different plane Burfaees. 

Ex. 2, Four points, not all in the same plane determine how 
many different places ? how many different straight lines ? 

Ex. S. Three parallel straight lines, not in the same plans, deter- 
mine how many diHerent planes ? 

Ex. 4. Four parallel straight lines can determine how many difter- 



Ex, 5. Two iiilerseetins straight lines an.! a point, not in thei!,' 
plane, determim; how many difie-rent planes 1 



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322 BOOK VI. tiorjD geometky 

Propositio>- I. Theokem 

509. If a siraiglht line is perpendicular io each of ttvs 
other straight lines at their point of intersection, it is per- 
pendicular to the plane of those lines. 




Given AB J. lines BC and BD, and the plane UN pass- 
ing through BC and BD. 

To prove AB ± plane MN. 

Proof. Through B draw BG, any other line in the plane 
MX. 

Draw any couveuieut line CD interseeting BC, BG and 
BB in the points C, G and B, respectively. 
Produce the line AB to F, making BF^^AB. 
Connect the points C, G, D with A, and also with F. 
Then, in the A ACD and FCD, CD=CD. Ident. 

AC- CF, and AD^BF. Art. 112. 

.-. AACB^AFCB. (Whyf) 

.-. / ACB= I FCD. (Why!) 

Then, in the A ACG and FCG, G6=CG, (WbyT) 

AC^CF, and / ACG = IFCG. (Why!) 

.-. A AGG^AFCG. (Whyt) 



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LINES AND PLANES 323 

.-. AO=GF. (Why?) 

. B and 6 are each equidistant from the points A and F. 

:. BO is X AF; that is, AB ± BG. Art. ii3. 

.-. AB X plane jl/iV, Art. 505. 

(forii is X any line, BG, in the plane MJS, through its fool). 

q. E. D. 



PROrOSITION IT. TlIEOItKM 

510. AU tiie perpeitdieulars that can be tirawn to a 
given line at a given point in the line lie in a plane perpen- 
dicular to the line at tJte given point. 



X 



■<ri\. 



Given the plane M]V and the line 5(7 both X line JBa,i 
the point B. 

To prove that BG lies in the plane MI^. 

Proof. Pass a plane AF through the intersectinff lines 

AB and BC. Art. 503. 

This plane will intersect the plane MN in a straight 

line BF. Ari. 508, 2. 

Bat AB X plane MN (Hjp.) .". AB X BF. Aft. 505. 

Also AB X BC. Hyp. 

.-. in the plane AF, BC and BF X AB at B. 

/. BG and BF coincide. Art. 71. 

But BF is in the plane MN. 

.: BC must be in the plane 31N, 
{for B(J uMWiklcs wUh BF, ichidi. ties m the jilaiic MX). 

e, E. B. 



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824 BOOK VT. SOTJD GFXl^ilETliY 

611. CoE. 1, At a given point Bin iliestm'ijhl line AB, 
io construct a plane, perpendicular to the line AB. Pass a 
plane A-F through AB in any convenient diveiiliou, and iq 
the plane AF at the point B coiiBtruet BF ± AB (Art. 
274). Pass another plane through AB, and in it construct 
BP X AB. Through the lines 
BF and BP pass the plane 
MX (Art. 503). MX is the f 

required plane (Art, 509). 



512. Cor. 2. Through a 
given external point, P, to pnsa 
a plane perpendicular to a 

given line, AB. Pass a plane through AB and P (Art. 
503) , and in this plane draw PB ± AB (Art. 273) . Pass 
another plane through A B, as AF, and in AF draw BF X 
AB at B (Art. 274). Pass a plane through BP and BF 
(Art. 503). This will he the plane required (Art. 509). 

513, Cor. 3. Through a given point hut one plane can 
be passed perpendicular to a given line. 



Ex. 1. Five points, no tour of which are in the same plaoe, deter- 
line how many different planes ! how many different straight lines 1 



A atvaight line and two points, not 
le, determine how macy difiecent plat 



Ex. S. In tte figure 
GDF are equal. 



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LINES AND PLANES 



Proposition III. Problem 



514. At a given point m a plane, to erect a perpendicu- 
lar to the plane. 



1 ''• 1 


sz^^^ 


\-^v i\ 





Given the point A in piano MN. 

To construct a line pei-pendicular to MN at the point A. 

Construction. Through the point A draw any line CD in 
the plane MF. 

Also through the point A pass the plane PQ X CD 
(Art. 511), intersecting the plane J/jV iu the lineiSS. Art, 508, 2. 

In the plane FQ draw AK 1 line ES at A. Art. 274. 

Then AK is thi; ± requirud. 

Proof. CD ± plane Py. Coiistr. 

;. CD 1 ,4/t:. Art. 505. 

Henw AK X CD. 

But il/r X 7i\S\ Coustr. 

.-. AK X phine MN'. Art. 5oy. 

Q. E. F. 

615. CoK. .(l( fl given point in a plane but one perpen- 
dicular to the plane can he drawn. For, if two X could be 
drawn at tlie given point, a plane conld be passed through 
them intersecting the given plane. Then the two X would 
be in the new plane and X to the same line (the line of 
intersection of llie two planes, Art. 505); which is im- 
possible (Art. 71). 



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■rlh BOOK VI. fiOLID CEOJIETilY 

Proposition IV. Problem 

516. Fimn a given point iviiliout a jilatie, to draic a line 
pcrpfndicifhii- to the plane. 




Given the plane MX and the point A external to it. 
To construct from A a line ± plane MX. 
Construction. In the plane MX draw any convenient 
iine BG. Pass a plane through BC and A (Art. 503), and 
111 this plane draw AD ± BG. Art. 273. 

In the plane MX draw LD X BC. Art. 274. 

Pass a plane through AD and LD (Art. 503), and in 
that plane draw AL A. LD. Art. 273, 

Then AL is the ± required. 

Proof. Take any point C in BG except X>, and draw LO 
uud A G. 

Then &. ADG, ADD and LDG are right A . Consti 

.-. J6''=Zd^ + DG'. Art. 400. 

.-. AC^ = ZL^ + LD' -f 'DC'. Art. 400, Ax. 8. 
.'. AC^^AL^ -{■ LG'^. Art, 400, As. 8. 

.-. ZAiCisaright /. . Art. 351, 

But AL ± LD. Conatr. 

.-. AL X MX. Art 



517. Cor. Bitt one perpendicular can be dratcn from a 
given external point ia a given plane, 



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LINES AND PLANES 327 

Proposition V. Theorem 

518. I. Oblique lines drawn from a point !o a plane, 
meeting the plane at equal distances from the foot of the 
perpendicular, are equal; 

II. Of ftco oblique lines drawn from a point to a plane, 
but meeting the plane at unequal dlniances from the fool of 
the perpendicular, the more remote is the greater. 




Given AB ± plane MN, BD=BC, and BH > EC. 
To prove AD^AC, and AM > AC. 
Proof. I. In the right A ABD and ABO, 

AB = AB, and BD = BC. (Why?) 

.-. AABZ>=A^£0. (Why') 

.-. AI> = AC. (Wby?) 

II. On Offtake iSJP^Be, and draw AF. 
Then AF^AC (lypart of theorrm jin^t prm'f,d) . 
But AIT > AF. (Why?) 

.-. AU > AC. As. 8. 

Q. E. D. 

519. Cob. 1. Conversely: Equal oblique lines drawn 
from a point to a plane meet the plane at equal distances 
from the foot of the perpendicular drutni from the same point 
to the plane; and, of two unequal li}iex so drawn , the greater 
line meets the plane at the greater distance from the foot of 
the 'perpendicular. 



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328 



SOLID r.EOJlETl'.Y 



520. r'dJi 2. The locus of a point in space cijuklhiant 
from all llif points in the circumference of a circle is it 
ulraight line passing through the center of the circle and 
perpendicnhtr to its plane. 

521. Cor. 3. The perpendicular is the shorlpst line Ihut 
can be drawn from a gii-en point to a given plane. 

522. ]>KF. The distance from a point tu a plane is the 
perpendleiihir drawn from the point to the jilniit;. 

Proposition VI. Theorem 

523. If from the foot of a perpendicular to a plane a 
line be drau-n at right angles to any li7te in the plane, the line 
drawn from thtpoint of intersection so formed to any point in 
the perpendicular, is perpendicular to the line of the plane. 




Given A7i ± plane MX. and BF X CJ), nny Hue in 3LY. 
To prove AF J_ CJJ. 

Proof. On €1) take FP and FQ equal segments. 
Draw A P, HP, AQ, BQ. 

Then BP^BQ. Art. 112. 

Hence AP^AQ. Art. ma. 

.'. io the line AF, the point A is equidistant from P 
and Q, and F is equidistant from P and Q. 

.: AF ± €T). (Why!) 

Q. E. ». 

Ejc, In the above figure, it Ali^G, AF^S, and jy = lU, find QF, 
DF and C^. 



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LIKES AND I'LASES 



Proposition VII. Theorem 

524. Two straight lines perpetnUcitlar to ike same plane 
>■€ parallel. 





i 

\ 




/ 


\ 


•f / 


/ """"T I 


Given the lines All ami CD ± plane ilW. 


To prove AB 11 CD. 


Proo(. Draw BI), anil Uu-ousli J), in the plane MS, 


draw FR X BI>. 


Draw Ai). 


Then 


;j/< J- FII. Con.lr 



Ai> ± FM. Art. 023. 

0/' i F7Z. Art. 505. 

.-. BB, AD and (7/") rn-p all ± J'J/ at the poiut T>. 

.". R/', A7' and Cf.) all lie in the same plane. Art. 5io. 

.-. AB imd CD ave iu the same plan'^. (Why?) 

But Ali and (JIf are ± B'li. Art. 50o. 

.-. AB and 6'Z> are 1|, Art. vi\. 

t). E. D. 

525. Cor. 1. If one of tiro parallel liiie^ j.v pn-pendicn- 
iar to a picnic, the other is perpendicular to the plane also. 

For. if AB and CI) be ||, and AB ± 
plane PQ, a line drawn from C L TQ 4 f 

must be II AB. Art- ;i?4. 

But CP iiuist coincide with tin? line V ^ ;)\ 
so drawn (Art. 47, 3); .'. CD 1 7'(^ ^ —''^^ 



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:^,:";o 



SOLID GEOMKTP.V 



526. Cor. 2. Ifin-o straight lines are each parallel to 
a third straight Uhp, tJiey are parallel to each other. For, 
if a plane be drawn 1 tu llii? 1;liird line, each of the two 
other Sines must be 1. to it {Art. 525), and therefore be 1| 
to each other (Art. 5-4). 

y*iioi'OfiTio\ YIII, Theorem 

527. If a sirairjhi line ex/erna! to a given plane is paral- 
lel lo a line in the plane, then ihejirst line is parallel to the 
given plane. 




B Cn in the plane MN. 



!■ .¥_V. 



Given tlio straight linpyl/; II 

To prove Ml \\ plai 

Proof. Pass a plane thron^h the || lines AB and CD. 

If AB meets MN it must meet it in the line CT). 

But AB and CD cannot meet, for they are ||. Art, 120, 

.". AB and MN cannot meet and are parallel. Art. 506, 

Q. E. 9. 



528. Cor. 1. // a straight line is parallel to a plane, 
the intersection of the plane with any plane passing through 
the given line is parallel to the given line. 



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LINES AND PLANER 



529. Cor. 2. Through a given line ^_ ,g 

(CD) to pass a plane parallel to another q 

given line (AB). q..^i^^,^. r 

Through P, any point in CD, draw ^ 

QR |[ AB (Art. 279). Througli CDaui 
QB pass a plane (Art. 503). Tliis will be tlie plane 
reqaired (Art, 527). 

Jt AB and CD are uot parallel, but one pluiie can be 
drawn through CD\\AB. 



Proposition IX. Theorem 

530. Two planes perpendicular to the sanu 
re parallel. 



Given the planes MN and PQ X line A 11. 
To prove MN \] PQ. 

Proof. If MN and PQ a.ve not parallel, on being pro- 
duced they will meet. 

We shall then have two planes drawn from a point per- 
pendicular to a given line, which is impossible. Art. 5i3. 
.■, MN and PQ are parallel. Art. 507. 



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f;oLin (";En:MET];v 



Proposition X. Thk(jr!:m 



631. // two parallel planes are cut by a tJih-d plane, ike 
intersections are parallel lines. 




Given M'N' and PQ two II planes iiitersecteil by the planuj 
BS in tbe lines AB and CD. 

To prove AB \\ CI). 

Proof. AB and CT) lie in the same plane US. 
Also AB and CI) panKOt meet; for if they did meet 
the planes ilO' and FQ would meet, which is impossible. 

Art- r>07. 
.-. AB and ('/> are pm-illoi. Art. 4L 

0- E. D. 

532. Cor. 1. Parallel lines included between parallel 
planes are equal. For, if AC and BD are two parallel lines, 
a plane may be passed through them (Art. 503), intersect- 
ing MN and BQ in the 11 lines AB and CD. Art. 531. 
.". ABDC is a pariillelogram. Art. U7. 
.-. .1(7= CD, Art. 155. 



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LISES ASD PLANES 333 

533. Cor. 2. Two parallel planes are everywhere equi- 
distani . 

For lines 1 to oae of them are il (Art. 524). Hence 
the segments of these lines included between the || planes 
are equal (Art. 532). 



Proposition XI. Theorem 

534. If two infersecting lines are each parallel to a given 
plane, the plane of these lines is parallel to tJie given plane. 



\ : 


> 


1^ \. 








'\ 


,.■-'1 


"■J \ 



Given the lines AB and Cl>. eatih \\ plane PQ, and inter- 
secting in the point F: and lilN a plane through AB and 
CD, 

To prove MX \\ PQ. 

Proof. From the point F draw FS X PQ. 
Pass a plane through FC and Fff, intersecting PQ in 
BK; also pass a plane through FB and FE, intersecting 
PQ in EL, 

Then EK \\ FC, &nd BLW FB. Art. 538. 

But FE 1 EK and EL. Ait. 505, 

.-. FE 1 FC ;.iid FB. Art. 123. 

.-. FE ± MX. Art. 509. 

.-. MN WPQ. Art, ,^30. 



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d'ii BOOK VI. SOLID GEOJIEiny 

Proposition XII. Theoresi 

535. A straight line perpendicular to one of two parallel 
planes is perpendicular to the other also. 



\  


-- 


L_4 - \j 








\. 


.^••J 


-1 \ 



Given the plane MN || plane PQ, and AB L FQ. 

To prove AB L MN. 

Proof. Througrb AB pass a plane intersecting PQ and 
MN in the lines BC and AF, respectively; also throngh 
AB pass another plane intersecting PQ and MN in BJ) and 
AH, respectively. 

Then BC 11 AF. and BD \\ AS. 

But AB X BC and BB. 

:. AB 1 ^J'and^fl'. 
.-. AB X plane J/A". 

586. COK. 1. Through a given point to pass a plane 
parallel to a given plane. 

Let the pupil supply the construction. 

537. Cor. 2, Through a ginen point but one plane can 
be passed parallel to a given plane. 



Art 


531. 


Art 


505. 


Art 


123. 


Art 


5CB. 


0. S. 


>. 



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LIMES AND PLANEiS 335 

Proposition XIII. Theorem 

588. If two angles not m the same plane have their 
corresponding sides parallel and extending in the same direc' 
Hon, tht angles are equal and their planes are parallel. 



Y 



Given the Z.BAC m the plane MX, and the ZB'A'G' 
in the plane PQ; AB and A'B' \\ iind csteuding in the 
same direction; and AC and A'C W and extending in the 
same direction. 

To prove Z^^iO = Z B'A'C, and plane ilfiV |1 plane P^. 
Proof. Take AB = A'B\ and AC=A'C'. 
Draw AA', BB', CC, BC, B'C. 

Then ABB' A' is a ZI7 , Art. 160, 

(foi-ABandA'B'are^and [\). 
:. BB' and AA' are = and ll. Art, 155. 

In like manner CG' and AA' are = and !|. 

.-. BB' and CC are^and ||. (Why ?) 

.-. BGC'B' is a C7 , and BC=B'C'. {Why V) 

.-. A ABC= A A'B'C. (Why 5) 

.-. ^A = ZA', (Why?) 

Also AB II A'B', ,■. AB \\ plane PQ. Art. 527. 

Similarly AC 11 plane P(?. 

.-. plane MN || plane PQ. Art. 534. 



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doO BOOK V!, HOLID GEOMKTKY 

Propositiux XIV. Theorem 

539. // two straight Ihies are hilers€cletJ bf/ tlin^r purah 
lei planes, the corresyoiuVuuj svijairnia »/ tlic-'in lines are 
proportional. 



\ 



Given the straight iities AB and CD intersected by the 
1! planes MN, PQ and SS in the points A, F, B, and 0. 
H, D, respectively. 

AF CR 

To prove — 



: the plane PQ in 0. 



FB MB 
Proof. Draw the line AB Interseetiu 
Draw i^G. BB, GH, AO. 
Then FG \\ BB, B.\id. GR\\ AG. Art, 531, 

.■^'=^. Art. 317. 

FB GD 

And ^=f|. (W.,n 

•■ FB SB ' y'l 

Q. E. B. 

Ex. 1. la above figure, if AF=1, FB = 5, and CH^S, lind CD. 
Ex 2- If Cif=3, SD=4, and JB=10, find JJ-' and fif. 



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DIHEDRAL ANGLES 



DIHEDRAL ANGLES 

540. A dihedral aagie is the opening between two in- 
tersecting planes. 

Prom certain points of view, a dilitdriii angle may be regarded as 
a wedge or slice of space cut out by the planes forming the dihedral 
angle. 

541. The faces of a dihedral angle aro the planes form- 
ing the diliedral angle. 

The edge of a dihedral angle is the straight line in which 
the faces intersect. p 

542. Naming dihedral an- 
gles. A dihedral angle may bo 
named, or denoted, by naming 
its edge, as the dihedral angle 
-1J5; or by naming four points, 
two on the edge and one on 
each face, those on the edge 
coming between the points on 
the faces, as F^AB-Q. The 
latter method is necessary in naming two or more dihedral 
angles which have a common edge. 

543. Equal dihedral angles are diliedral angles which 
<ian be made to coincide, 

644, Adjacent dihedral angles are dihedral angles liav- 
ing the same edge and a face between them in common. 

545. Vertical dihedral angles are two dihedral angles 
having the same edge, and the faces of one the prolonga- 
tions of the faces in the other. 

548. A right dihedj'al angle is one of two c((iial adja- 
cent dihedral iuiglcs t'onued by two ])lanea. 




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338 



vy. 




547. A plane per- 
pendicular to a givoii 
plane is a plaiii; fovra- ^____„ 
ing: a right diliedra! \ 
angle with the givuu \ 
plane . \ 

Many of the properties of dihedral angles are obtained 
most conveniently by using a plane angle to represent the 
dihedral angle. 

548. The plane angle of a dihedral an- 
gle is the angle formed by two lines drawn 
one in each face, perpendicular to the edge 
at the same point. 

Thus, in the dihedral angle C-AB-F, 
if FQ is a line in the face AD perpendicu- 
lar to the edge AB at P, and PR is a line 
in face AF perpendicular to the edge AB 
at F, the angle QPIi is tiie plane angle of the dihedral 
angle C-AB-F. 

549. Property of plane angles of a dihedral angle. 

The magtiltmU of tlio plane angle of a dihedral angle is 
the same at every point of the edge. For let EAG he the 
plane I of the dihedral I E-AB-I) at the point A. 
Then PR || AE, and PQ || AC (Art. ll^l.) 
.-. IRl'Q - Z.EAC (Art. 538). 

550. The projection of a point upon a plane is the foot of 
a perpendicular drawn from tlie point to the plane. 

551. The projectioa of a line 
upon a plane is the locus of the pro- m 
jections of all the points of the line 
on the plane. Thus A'B' is the 
projection oi AB aa. the plane 31N. 




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DIHF.DIUL AXGLES 60)} 

Proposition XV. Tiieoreji 

552. Tico dihedral angles are equal if their plane angles 
are equal. 



lU .LI I 



Given Z DBF t! . | Iiik- I of the dihedral Z ChAB-F, 
IB'B'F' the plaiiL. ^ uf the dihedral ^C'-A'B'-F, aud 
ZDBF==Z.I)'B'F'. 

To prove Z C-AB-F = I C'-A'Ii'-F. 

Proof. Apply the dihedral IC'-A'B'-F to IC-AB-F 
so that Z li'Ii'F coincides with its equal, IDBF, 

Gaom. As. 2. 
Then line A'B' mirst coincide with ,4.ii, Art. 515. 

{fof A'B' and AB are both X jAu.eiJBFut ihepoiiiIJi). 

Hence the plane A'B'D' will coiueido with plane ABD, 
Art. 501. 
(ihroiigh tao taUmecting lines uiih/ one pliiiie mit he [tti^seil). 

Also the plane A'H'P will coincide with the plane ABF, 

.". Z (7'-A'/?"-7'"" coincides with Z 6'--4/>'-i'' and is equal 
to it. Avt. tr. 



558. Cor. The vertical dihedral oni/I''^ Jhnu'-d h 11 tin) 
intersecting planes are equal. 

In like manner, many otlicr [n'opertii^i^ of phuie aiitrlis 
are true of dihedral angles. 



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340 )!noK VI. SOLID 'ieoi]i:jrv 

Proposition XYL Theorkm 
554. Tu-o dihedral aiuih^ li'iv' thr .s^imc r 




.■-■,^B 



Given Uie Jihe.lral A V-All-D rikI C'-AT,'-!)' liaviug 
the plane A fMZ) and C'A'T*', respectively. 

To prove I C'-A'B'-n' -. lC-Ali-1) =IC'A'!Y -. 
I CAB. 

Case I. When the plane A <"A'J>' ami CAT) (Pigs. 2 
andl), are comtneiisurahk'. 

Proof. Find a eommou measure of tlie A V'A'D' aiid 
CAD, as Z GAK, and let it be contained in Z G'A'D' n 
times, and in LGAT) m times. 

Then / C'A'iy : I CAD^n -. m. 

Through A'B' and the lines of division oE Z C'A'D' pass 
planes, and through Ali and the lines of division of 
Z CAD pass planes. These planes will divide the dihedral 
IC'-A'B>-D' into n, and ZC-AIi-D into m parts, all 
eqnal. Art, 5S2. 

.-. ZC'-A'B'-D' : lC-AB-I)=n -. y,i. 

Hence Z C'-xVB'-D' : Z C-AB-7J = Z C'A'D' -. Z CAD. 

(Whj ?) 



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DlHF-DliAL ANGLES J41 

Case II. When (he plane angles C'A'D' and CAB (Figs. 
3 and 1) are incommensurable. 

Proof. Divide the -Z CAD into any miraber of equal 
parts, and apply one of these parts to the I C'A'D' . It 
will be eontiiined a i;ertain nnmber of times with a remain- 
der, as ^LA'D', less than the unit of measnre. 

Hence the i G'A'L and GAD are commensurable. 
.-. I C- A'B' ~r,:l C-AB-D^ / fJ'A'L : Z CAD. Case I- 
If now we let tho unit of measure be iudefiiiitulv iliniiii- 
iahed, the Zi.-l'/)', which is less tliau the unit of measure, 
wiil be indefinitely diminished. 

.-. I CA'L i I C'A'D' as a limit, and 

iC'-A'ir-L^ lC'-A'B'~D' ■i^s-^WmM. An. 2&L 
-A'B'-L 



Hence 
Z C'A'B'-D' 



IC-AB-D 



becomes a variable, 



vith 



limit. 

But the \ 



tnes a variable with 



^C'-A'B'- L _ 



Z.C- Mi-B 




I C'-A'B'-B' 



Ex. I. How many straifjiit lines nre neeessafy to indicate a dihe- 
dral nngle (as IE-AB~II, p. 338)? IIow many straight lines are 
neeesBary to indieate the plaoe angle of a dihedral angle 1 Hence, 
■what is the advantage of naing a plane angle of a dihedral aogle 
instead of the dihedral anf;le itfielE f 



Ex 2, 

ogoug to p: 



B thre 






iial propert; 



[ dihedral angles anal- 



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Proposition XVII. Theorem 



555. If a tstrhUjkt line is pn-pendic\dar to a plane, every 
plane ilmini fliroiigh iluil line in pcrpemUcular to the plane. 




Glvea the line Ali X plane MN, and the plane PQ 
through A B and interseethig MN in EQ. 



To prove PQ X MX. 

Proof. Ill the plane .'l/.Vdi-aw BO ± RQ at B. 

Bnt AB X JiQ. Art. 505. 

.-. ^ABC U the plane / of the dihedral ZP-RQ-M. 
Art. 548. 
But /.ABC is. a visht I , Art. 505. 

(for All 1 MNhy li-ip.). 



556. Cor. A plane perpendicular f" the ed<je of a 
dihedral angle is perpendicular to eacli of the two faces form- 
ing the dihedral aixfjle. 



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dihedral anglbs 343 

Proposition XVIII. Theorem 

557. If two planes are perpendicular to each oilier, a 
straight line drawn in one of them perpendicular to their 
line of intersection is perpendicular to Ike other plane. 



Given the plane PQ X plane 3fy and intersecting it in 
the line RQ; and AB a line in PQ X HQ. 
To prove AB X plane MN. 

Proof. In the plane US' draw BC X EQ. 
:. ZABCis the plane I of the dihedral ZP-RQ-M. 
Art. 548. 
.-. Z^ECisart. /. . Art. 554. 

{/hi- I'-EQ-Mu a ri'jht <lihe<lna I ). 
.■, AB X BO and fi(^ at their intersection. 

.-. AB X plane WN. (Why %) 

Q. £. 9. 

558, Cor. 1, If two planes are perpendictiiar to each 
other, a perpendicular to one of them at any point of their 
intersection will lie in the other plane. 

For, in the above figure, a X erected at tlie point B in 
the plane MX must coincide with AB lying In the plane 
PQ and X MN, for at a given point in a plane only one X 
can 1)6 drawn to that plane (Art. Slij). 

559. CcR. 2. If two planes are perpendicular to each 
other, a perpendicular to one "plane, from a point in the other 
plane, will lie in the ollu-r plane. 



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DOOK Vf. ^OLID GEOMl^TRY 



PKOPOSITION XIX, THEOJiEM 



560. // iu'O inUrsecHntf plni 
a third plane, iheh- Unr of ■in 
the third plane. 



', each pBrppnd'cnUir 
ion is iKvpi-'udicKtar 




Given the planes PQ and BS 1 plane Mlf, and inter- 
secting in the line AB. 

To prove AB ± plane MX. 

Proof. At the point B in wliich the three planes meet 
erect a ± to the plane MX. This ± must He m the plane PQ. 
and also in the plane RS. ^'^^- ^'^'^■ 

Hence this L must coincide with AB, the iiitersfction 

of PQ and RS. Art. 50S, 2. 

.-. AB ± plane .1/-V. 

g. E. B. 

561. Cos. If two planen, indmVnig a right dihfdra! 
angle, are each perpendicular to a third plane, the inferspc- 
tion of any two of the planes is perpendicular to the third 
plane, and each of the three lines of intersection is perpen- 
dicular to the other lu'o. 



Ex. 1. X»me all iha dihaiiral an^ 
Ez. 2. If Z CBQ^SO", find the ri 



a on the above fijjiire. 

o£ ea.eli pair of diliedral d. 



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DIHEDRAL ANGLES 



Proposition XX. Theorem 

562. Every point in the plane which iiseets a given 
dihedral angle is equidistant from the faces of the dikfdral 

angU- 




Given plane CB bisectins? tlie diluidi-al I A-BR-T), P 
any point in phme ISC, I'Q iiud FT 1. faces BA and IW, 
respectively. 

To prove I'Q^PT. 

Proof. Through PQ and FT pass a plane interseetmg 
AB in QE, BD in KT.and BG in Pli. 

Then plane FQT J. planes AB and BD. Art. 555. 

.■, plane PQT ± line RB, the intersection of the planes 

AB and BD. Art. 5fi0. 

.-. RB L RQ, EP and BT. Art. r.or.. 

.-. i QBP and P/C7' ni^e the plane A of tht dihedral 



A A-BR-P and F-BE-D. 

But these dihedi-al ii are orinal. 

.-. LQRV= IPRT. 

:. rt.A P()/^ = rt, A PRT. 

:. PQ^FT. 



Art. 54S. 

Hyp. 



563. n^ tonisof all points 
of a (lilu'dral angle is the plane t 



(Why?) 
Q. E. ». 

■qni<li.slant from- Ihe fares 
^r:ciiitg ihv ilihi'iirul angle. 



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1U10K VI. SOLEII 



XXI. PiiOBLGM 



564. Through unij Ktrdi'jht I'li". not perpendicular to < 
giten plane, to pass a plane pcriKiidkular to the given plane 




Given the line AB not ± plane MN. 

To construct a plane passing through AE and ± MN. 

Construction. From a point -1 In ths liue AB draw a 
AC to the plane MN. 

Throngh the intersecting lines AB and AG 
plane AD. 

Then AB is the plane required. 

Proof. The plane AD passes throngh AB. 

Also plane AD ± plane MN, 



{M 



sAC, ichichig 1 MN). 



Art 


51(1. 


pass 


tlie 


Art 


503. 


Conefr. 


Art 


555. 


Q. E. 


r. 



565. Cor. 1. Through a struigJd line not perpendicular 
to a given plane only one plane can be pmsed perpendicu- 
lar to thai platie. 

For, if two planes could be passed through AB ± plane 
My, this intersection AB would be X MN (Art. 560), 
which is contrary to the hypothesis. 

566. COE. 2. The projection upon a plane of a straigM 
line not perpendicular to-that plane is a straight line. 

For, if a plane be passed through the given line X to 
the given plane, the foot of a X from any point in the 
line to the given plane will be in the intersection of the 
two planes (Art. 5oy), 



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DIHEDRAL ANGLES 347 

PROPOsiTiON XXII. Theorem 

587. The acute ant/le which a line maki's ivith Us pro- 
jeetion on a plane is the least angle which it makes with aiiy 
line of the plane through its foot. 




Given line AB meeting tlie pianp MN in the point B, 
■BC the projection of AB on MN, and PJi any other line 
ia the plane MN through B. 

To prove that ZIBC is less than ZAIiP. 
Proof. Lay off PB equal to CB, and draw AG and AP. 
Then, in the A ABC and ABP, 
AB=AB. 
BG^ BF. 
Ent AG < AP. 

.: Z^BC is less than lABP 



(Why t) 



Art. 108. 

Q. E.D. 

568. Def. The inclination of a ihie to a plane is the 
'ciite angle which the tjiven line makes with its pi'ojsetion 
'Pon the given jjlaiie. 

E». 1. A plane has an inelination of 47" to ea«tt o£ the ta^ne of a 
iliadral angle and is parallel to clie edge of tbe dilieiiral angle; how 
aonj degrees are in the pkne UHglt of the dihedral an^'l..'; 

Ex. 2. In llie fiL-iire ™ pae? 3-ir,. if PT=QT,liow large lathe 
'''"dral z A -!:il-li'i i[ n---llT, hmv large Is it? 



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SOT.ID gi:u.mk-i'i;y 



Proposition XXIII. Problem 



569. To <lruw a common pTpeu'/'Cidar to any tivo Hi, 
tiof in the same plane. 




Given the lines AB and VJJ not in the same plane.' 
To construct a line perpenLliculiir to botli AB and CD. 
Construction. Tlirongh AB i>i(..3=! a. plane MXW line CIk 

Art. j>! 

Through CD pi^s a piano CF 1 ])laiie M^' (Art. 564), aui 
interaecting plane MN in the line BF. 

Then EF \\ CD (Art. 528), .-. EF must intersect Ai 
(which is not |1 CD by hyp.) in some point K. 

At E in the plane CF draw LK ± EF. Art. 274 

Then LK is the perpendicular reqitired. 

Proof. LK ± EF. Con^ir 

.-. LK ± CD. Art. l:.;. 

Also LK ± piano MX. Art. ss: 

.-. LK _L line _l7i. (Wl,y^ 

.-. LK ± both C« and AB. 

Q. E. ?. 

570. Only one perpendicular can be draicn between ttco 
lines not in the same plane. 

For, if possible, in the above figure let another Hue BD 
be drawn ± AB and CD. Then, if a line be drawn throagh 
B II CD, BD X this line (Art. t2;t), and .-. X plane MN 
(Art. 509). Draw DF X Hue EF; then Z>F X plane MS 
(Art. 557), Henoe from the point T) two X, DB and DF, 
are drawn to the plane MX, which is imposBiblu (Art. 517), 



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rOLYIIEDUAL Af^dLK 




POLYHEDRAL ANGLES 

571. A polyhedral angle is the amount 
o£ opening between tlireu or more planes 
meeting at a point. 

Such an angle may be regarded as a portion of 
spBoe cut out by the planes torming the anfila. 

572. The vertex of a polj-hedral angle 

is the point in which tlie planes forming the angle meet ; the 
edges are the lines in which the pianes intersect; the faces 
are the portions of the planes forming the polj"lie(ival angle 
which are inclnJed between tlie edges ; tlie face angles are 
the angles Eormeil by the edget;. 

Each two adjacent fae.es of a polyliedral angle form a 
(iihedral angle. 

The parts of a polyhedral angle ;ire its faee angles ami 
liihedval angles taken togetlier. 

573. Naming a polyhedral angle. A polyhedral angle 
ia named eitlier by namiuij the vertex, as I': or by naming 
the vertex and a point on each edge, as V-ABG. 

In case two or more polyhedral angles have the same 
Vertex, the latter method is necessary. 

In the above polyhedral angle, the vertex is Y; the 
edges are VA, YB. T"C; the face angles 
are J.rjS, BVC, AVC. 

574. A convex polyhedral angle is ii 
polyhedral angle in which a suction 
ttiade by a plane cutting all the edges 
IS a convex polygon, as y-ABCI>E . 

575. A trihedral angle is a polyhe- 

iral angle having tliree t'aces; a tetrahedral angle is one 
laving four face*, etc. 




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GEOJIETliY 



576. A trihedral angle is rectangular, birectangular, or 
trirectaogular, aeeordinK as it contains one, two, or /ht-ee 
right dihedral angles. 



577. An isosceles trihedral angle i 
two of whose faee angles are equal. 



trihcdi-al angle 



578. Vertical polyhedral angles an 
having the same vertex and the faees ( 
the other produced, 

579. Two equal polyhe- 
dral angles are polyhedral an- 
gies bavin;^ their correspond- 
ing parts equal and arranged 
in the same order, as V-ABC 
and V~A'B'C. 

Two equal polyhedral angles 
may be made to coincide. 

580. Two symmetrical poly- 
hedral angles are polyhedral an- 
gles having their corresponding 
parts equal but arranged in 
reverse order. 

If the faces of & trihedral angle, V-ABC, 
produced, they will form a vertical trihedral angle, 
V-A'B'C, which is symmetrical to V-ABC. For, 
if V-A'B'C be rotated forward about a horir.ontal 
aiis through V, the two trihedral angles ai 
to have their corresponding parts equal I 
ranged in reverse order. 

Similarly, any two verlieal polyhedral 
are aymmetrieai. 



polyhedral angles 
: one the faces of 




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P0I.YI!E1>R.\L ANGLES 



351 



581. Equivalence of symmetrical polyhedral angles. It 
has been shown in Plane Geometiy (Art. 488) that two 
triangles (or polygons) symmetrical with respect to an 
axis have their corresponding parts equal and arranged 
in reverse order. By sliding 
two such figures about in a plane 
they cannot be made to coincide, 
but by lifting one of them up 
from the plane- in which it lies and tnrnii 
be made to coincide with the other figure. 

Symmetrical polyhedral angles, however, cannot be 
made to coincide in any way; henee some indirect method 
of showing their equivalence is necessary. See Ex. 29, 
p. 358. and Arts. 789-792. 



 it mav 



8 figUl 



Prop. XX. rf 
those OQ tlie figure ! 



Kr, 1. Name the trihedral angles 
i^PBQ^m", what kind of trihedral a 
If iLPBQ=m°, whot kind are they? 

Ei. 2. Are two trireot angular trih 
Prove this. 

Ex. 3. Are two lines which are perpendicular to the same plane 
necesaarily parallel f Are two planes which are pei'pendicular to the 
same plane necessarily parallel? Are two planes which are perpen- 
dicular to the same line necessarily parallel f 

Ek. 4. Let tlie pupil out out three pieces of pasteboard of the form 
ndioated in the aoeoropauying figures; cut them half through where 
'ho lines are dotted; fold them and fasten the edges so as to form 
hree trihedral angles, two of which (Figs. 1 and 2) shall be equal 
iEd two (Figs. 1 ;ind 3) sjmmetrical. hy esperiment, let the pupil 
ind which pair may be made to ooinuide, aud whifh not. 




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V[. SOLID GEOMIlTILV 



I'KOl'OfcJTIOX SXIV. TUEOKEM 

582. Thr sum of nnij fro f>w<' mujU, of a trlhedru! 
an'jtc is greater than the third faca atKjh. 




Given the trihedra! angle. S-ABC, with angle A8C its 
greatest face angle. 

To prove lASB + I HSC gvaai-n' {linn lASC. 
Proof. Ill the faee AS(^ ihviw n;>, itiakiiig ZASJ)^ 
ZASli. 

Take *7>=,S7>'. 

In the faee ASC draw the Hne AUC in any convenient 
direction, and draw AB and BV. 

Then, in the A A8B and ASB, SA^SA. 
SB = SD, and Z,l«fi= ZAS'I). 
:. A A8B=A ASVf. 
.-. AB=AD. 
Also AB + BC> AC. 

Hence, subtracting the equals AB and AJ). 

BC> BC. 

Hence, in the A BSC and DSC, 80=^ 8C, «B=Si>, and 

BC > DC. (Why!) 

.-. /. BSC is greater than Z DSC. Art. 108. 

To each of these uneqnals add the equals ZASB and 

ZASD. 

:. ZASB+ :/ B«0 is greater than I ASC. (Whyt) 

0. E- D. 

Ex. Iq the above figure, if IAj'<C equals one of the oilier £ace 
angles at S, as lASB, bow is the theorem proYod J 



(Why!) 
(Wl,y=) 
(my ?) 

(Why?) 
(\Vby?) 

(Why!) 



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POLYHEDKAL ANGLES o5d 

Proposition XSV. Theorem 

588. The sum of the face angles of any convex polyhedral 
angle is less than four right angles. 




Given i.l!c polyhedral angle S-ABCDE. 

To prove the sura of the face ^ at 5 less than 4 rt. ^ . 

Proof. Pass a plane cutting the edges of the given poly- 
hedral angle in the points A, B, C, J), E. 

Prom any point in the polygon ABODE draw OA. 
OB, OC, oh, OE. 

Denote the A having the common vertex S as the S A, 
atid those having the common vertex O as the A. 

Then the sum of the A of the S A=^ the sum of d of 
the A. AH, 11)4. 

Bat ZSR'l+/S7J0iagi-eatert!ian lABC, 1 , . „„ 
Z5C/J+ ZSCiMsgreaterthan ^BOT, etc.) ' 

•'. the sum of the base <i of the S A > the snra of the 

tase A of the A. Ajt, 9. 

.■. the sum of the vertex d of the S A < the sum of 

Uie vertex A of the A, As. ii. 

{if unequal s be suhtratilid from ctiiiiih, Ihe remainders are iiiicqiinl 
iiirermeor'Ui). 

But the sum of the d at 0=4 li. i . (Wiiy?) 

.'. the sum of face d at ii < 4 I't. A . Ax. s. 



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<i04 BOOK V!. SOLID GEOMKTRY 

PnOPOSITlOX XXVl. TlUXlREM 

584. JS iii-0 Irihedml ainjhs luu-c ihe thnc face migh^, 
of one eqmtl to the tliree fuce uiigl'-s of ilie other, ike (yi- 
hedral angles have their correspiinding dilwdml niifjUsequn}. 
and are- edlher equal or sijmmptriruJ . iircordiiir/ as fl/d,- 
corresponding face angles are arraniU'd in, f/w mine or Ui 
reverse order. 




Given thf trihedral i ^-ABC and ^'-A'B'C, liavin- 
the face A AUB, ASO 3.\\<\ BWC'equal to the face A A'ii'li'. 
A'S'C and B'S'C, respectively. 

To prove that the con-espon<3ing dihedral A of S~ABf' 
and S'-A'B'C are equal, and that A S-ABO and N'- 
A'B'C are either equal or symmetrical. 

Proof. On the edges of the trihedral A take ^S'.l, .S7>', 
8C, S'A', 8'B', S'C ail equal. 

Draw AB, AC, EC, A'B', A'C, B'C. 

Then, 1, In the A ASB and A'S'B', SA = S'A', SB = 

S'B', and AASB= ZA'S'B'. (Why?) 

.-. A ASB^A A'S'B'. (Why!) 

.-. AB^A'B'. (Why?) 

2. la like manner AO=A'C', and HC = B'C". 

:. A ABC^A A'B'C. fWhyf) 



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EXERCISES OS THE LINE AND PLAKE 355 

3. Take D a convenient point in SA, and draw DE in 
the face ASB, and DF in the face ASC, each ± SA. 

DE and i>F meet AB and .10 in points E and JP, 
respeetivelv. 

Similarly, take S'lf^SD and construct A J>'l-yr'. 

Then, in the rt. ^ ADE and A'B'E', Alt^A'l)', and 

Zi>jli7-./rA'l,". (Why?) 

.-. A ;inj;-A A'D'E'. (Why?) 

.-. A/^'=^'i", !ind !)E=I)'E'. (WliyM 

4. In like manner it m:\y\K sho^yu that AF=A'F', and 

.-. A .4Z:F=A J.'£'F'. (Wii;-?) 

And EF^E'F' (Whyf) 

5. Hence, in the A Z)i;F and B'E'F', T>E=n'E', I>F 
= D'F and EF=E'F'. (Why ?) 

.-. A DEF^A D'E'F'. (Why f) 

.-. IEJ}F=IF:J)'F. (Why?) 

Bat these i are the plane i of the dilieitral A whoso 
edges are S.l and N'A'. 

.-.dihedral Zif-Afr-C^diliedrnl IB'-A'S'-C Art. 5.12. 

In like manner it may be shown that the dihedral A at 
SB and S'B' are equal; and that those at KO and S'C are 
equal, 

.'. the trihedral A S and S' are either equal or sym- 
metrical. ^^'^^- 5'9' ''^o. 

Q. E. B. 
EXERCISES. CROUP 6't 

THEOREMS CONCEKNIXG TUE LINE A\D f'LANE IN SPACE 
Ex. 1. A segment of a liiie not parallel to a plane is longer thira 
its projection in the plane, 

Ex. 2. Equal stE-aight lines drawn from a point to a place are 
equally incliiit-a to the plane. 



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35G B 

Ex. 3. A line 
parallel. 

Ex. 4. If thre 
perpendicular to a 



SOt-in GEOMETRY 
H perpeudiculiir to the si 



ig in three stniight lines acs 
£ iulerscction Jii't parallel. 



Ejt. B. If a plane bisects any line at riglit 
angles, any point in the plane is equidistant 
from the ends of the line. 

Ex. 6. Given JIl X plane .If-V, 
and ,)(.' ± piano SS; 
prove JIC ± yii. 
Ex. 7. Given PQ X plane .1/-V, 
PR X plane HL, 
and BS 1 plane J/A'; 
prove QS X AB. 
Ex. 8. If a line is perpendicular to one 
of two intersecting planes,. its pro.jpctioQ 
on the other plane is perpeniJiQuliir to the 
line of intersection of the two planes. 
Ex. 9. Given CE X BE, 
AE I T>E, 
and Z C-AD-E a rt. dihedral Z 
prove CA X plane DAE. 

Ex. 10. The projections of two parallp 
a plane are parallel. (Is the converse 
theorem also true ?) 

Ex. 11. If two parallel planes are cut hy two non -parallel planes, 
the two lines of intersection in each of the parallel planes will make 
equal angles. 

Ex. 12. It a line is perpendicular to a plane, any plane parallel 
to the line is perpendicular to the plane. (Is the converse true !) 




Ex. 13. In the figi 
DC; prove BF X DC. 



Prop. VI, given AFi X MN and AF X 
X DC. 

Ex. 14. Two planes parallel to a third plane are parallel to each 
[Sua. Draw a line X third piano.] 



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EXERCISES ON THE LIKE AND PLASB 



Ei. 15. 
straight liii 


The 


projoetioiis up 
9 equal nnd p^r 


on a plane of t 

illel. 


Ex. 16. 

section. 


Al 


ne parallel to t 


vo pluues is p 


Ki. 17. 

prove the a 


la 
ngle 


the flguro to Py.p. XXII. it 
JRi' obtuse. 


Ei. 18. In a quadrilateral in apace (i. e., a 
quadrilateral whose vertices are not all in the 
same plane), show that the lines .ioliiing thu 
midpoints of the sides form u parallelogram. 


Ex. 19, 

the oppoait 


The 
esid 


lines joiniug the midpoints o 
s of a quadrilateral in Space Ij: 


Ex. 20. 

three faces 


The 


planes biseeti 
Hue every po 


g the ilihedral 
ut of wliii;h is 


[SUQ, 


Sbo. 


rt. m2.] 




Ex. 21 


7 


n 0(J bisecting I ROS, 
PQ X plane EOS, 
QR X OR, 
QS X OS; 
vo PIl=PS, PR X OR, 
□ii PS X OS. 



■qual and par 
pavallel to their ii 




Ex, 22. In n plane bisecting a piveu plane nn^^e, and perpendicu- 
lar to its plane, every point ia equidistant from the sides of the angle. 

[Sl'q. See Ei. 21 ; or through P any point in the bisecting plane 
pass pianes X to the aides of the iL , etc.] 

Ex. 23. In a trihedral angle, the three planea bisecting the three 
face angles at right angles to their respective planea, interaeet iu a line 
every point of which is equidistant from the three edges of the tri- 
hedral angle. 

Ex. 24. If two faceangleaof a trihedral augle are equal, the dihe- 
dral angles opposite them are equal. 

Ex, 25, In the figure to Prop, XXIV, provi) that lASC-'r IBSC 
ia greater than Z ASl) -{- Z ISSU. 

Ex. 26. The common perpendicular to two lines in spui^e is tho 
shortest line between thero. 



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'■^A 



vwr\ 



i;noi; vt, rood 

. Given .l/.Vll Jin, 
and A AIIC => A iibc. 



trihedral angles are equal. 

El. 29, Any two sjmmelrieiii trih, 
Aval angles are equivalent. 

[Suo. Take S.l, Sli, sr, S'a'. 
S'li', S'C', all fqiml. Pass planes -IfiC, 
A'B'C. Draw SO aiid S'ty ± tlieso 
plitnes. Tben thw trihedral A am 
divided into three pairs of isoseelos 
symtuetripal trihedral d , etc.] 



EXERCISES. CROUP 68 

LOCI IN SPACE 
Find the loeiis ol a point equidistant from 

Ex. 1. Two parallel planes. Ex. 3. Three t;ivea points. 

Ex. 2. Two given points. Ex. 4. Two interseoting linei 

Ex, 5. The three faces of a trihedral angle, 

Ex. 6. The three edges of a trihedral angle. 
Find the ioeus 

Ex. 7. Of all linos passing tlirongh a si^en point and parallel 




L plven plaice. 








Ex. 8, Of all liuPB pcrptiKiicul 


ar to a 


given line 


 It a given -point 


n the line. 








Ex. 9. Of all points in a give 


n plani 


3 eqnidisti 


mt from a given 


loint outside the plane. 








Ex. 10. Of all points equidista 


ut froii 


1 two £ivei 


1 points and from 


,wo parallel pianos. 








Ex. 11. Of all points equidista 


lit frou 


1 two give] 


11 points and from 


Swo intersecting planes. 








Es, 12. Of all poinis at a give 


■n dista 


iicu from 


a given plane aod 



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EXERCISES OK THE LINE AND PLANE 6Di} 

EXERCISES. CROUP 66 

riiOllLKMS CONCERNING THE POINT, LINE AND PLANE 

IN SPACE 
Ex. 1. Through a given point pass a plane parallel to a given 
plane. 

Ex. 2, Through a given point [lass a plane perpsniiii:iilai' to a 
given plane. 

Ex, 3. Tlirougii a given point to (construct a plani' paralli'l to two 
given lineR which are nol in the suiiie plai^f. 

I'rovf that only oui! piaue oun be coiistrurli^il fiilfimii!; Iho givi^n 
conditions, 

Ex. 4, Bisect a given iliheilral angle. 

Ex. 5. Draw a plane equallj inclined to three lines whleh meet at 
a point. 

Ex, 6. Through a piven point draw a line parallel to two givan 
intersecting planes. 

Ex. 7. Find a point in a plane .such that lines drawn to it from 
two given points without the plane make equal angles with the plane, 

[SuQ, See Ex. 23, p. 176.] 

Ex, 8, Find a point in a given line equidistant from two given 
points. 

Ex, 9. Fhui a point in a plane equidistant from three given points. 

Ex. 10, Find a point equidistant from four given points not in a 
plane, 

Ex. 11. Through a given point draw a line which shall intersect 

[Sua. Pass a, plana through the given point and one of the given 
lines, and pass another plane through the given point and the other 
given line, etc.] 

Ex. 12, Through a given point pass a piano cutting the edges of 
a tetrahedral angle so that tlie section shall be a parallelogram. 

fSVG. Produce each pair of opposite faces to interseet in a 
Straight line, etc.] 



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Book VII 
rOLYL{EDRONS 

585. A polyhedron is i ^lolid buiimlnl \i\ pUnes 

586. Tha faces o£ a pohlie 
6vou ar(i itri houmliiiy: pfiiiiLS tin, 
edges of a polyliedrun are tlii. hne% i 
of intersect ion of its faces 

A diagonal of a polyhedion lo i 
a straight line joining two ot ita 
vertices whicli are not m the j 
same face. The vertices of i polv- 
hedrou are the points in wliiji iti, 
edges meet or intersect. 

587. A convex polyhedron is a i>olylK'dvo!i in which a, 
section made hy any phme is a convex polygon. 

Only convex polyhedrons are to be considered in Ibis book. 

588. Classification of polyhedrons. Polyliedrons are 
simetimi'i elibsififd tttoidm^ to th mimlei of their 
fices Thus a tetrahedron it, a pohhedion ot foui fices 
T hexahedron is a pohh^dion of -ii^ fn. ^ in octahedron 
1 n rf tiglt I dodecahedron one if tftcl\c ml an 
cosabedron one ot twcntj f^ces 




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POLYHEDKOXS 




The polyhedronB moat important in practiea! life are those deter- 
mined by their stability, the facility with which thijy can be made out 
of eommoQ raateviuls, a3 wood and iron, tha readiiieaa with which 
they can be packed togethti', etc, Thus, pfism meaua "aomething 
Bawed oil." 

PEISMS AHD PARALLELOPIPBDS 

589. A prism is a polyhedron bounded 
bj two parullcl planes acd a group of 
planes whose Hues of iiitersectiou ui "" 
parallel. 

590. Tlie bases of a prism are tbe 
faces formed by tbe two parallel planet., 
tbe lateral faces are the faces formed h-\ 
the group of planes whose lines of inteiseetKm aie paiallel. 

The altitude of a prism is the perpendicular distance 
between the planes of its bases. 

The lateral area of a pi-ism is the sum of the areas of the 
lateral faces. 

591. Properties of a prism inferred immediately. 

1. The. lateral edges of a prism are equal, for they are 
parallel lines iucluded between parallel planes (Art. 589) 
and are therefore equal (Ai-t. 532). 

2. The lateral faces of a prism are parallelograms (Art. 
160), for their sides formed by the lateral edges are equal 
and parallel. 

3. The bases of a prism are equal polygons, for their 
homologous sides are equal and parallel, each to each, 
(being opposite sides of a parallelogram), and their liomoi- 
o?ous angles are equal (Art. 538). 



592. A right section of a 
plane perpendicular to thu hit' 



priM 



section made by a 



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3li2 1;00K VII. SOLID ceometisy 

C.O3 \ trianiFiilnr nri«m \< n iirUin ivlio';.' liilKf 



n I 



it 




594. All oblique prism is ii prism wliose Literal edges 
i-e ol)lir|iie to the liases. 



595. A right prism is a prism whose 
lateral {idgcs are pei'pemlieiitar to the bases. 



A regular prism is a right prism f 
ses are regiilai- polygons. 1 



whose bases 

597. A truncated prism is that part <if 
a pvisiii included between a base and a 
section nnide by a plane obliiine tu llie 
base and cutting all tlie lateral edges. 

598 \ parallelo piped j wl e 




ii II n r I'll ' " ' 



599. A right parallelepiped is a purallelopiped whose 
lateral edges are perpendicular to tiie bases, 



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PmSJtS AND rAKALLELOPIPEDS 363 

600. A rectangular parallelopiped is a right pavuUelo- 
piped whose bases are rectanglos. 

Hence, all the fitces of a rcciuiujidur puralldojyipcd are 
rectanglns . 

601. A cube is a rectangulai- parallelopiped whose edges 
are all equal. 

Hence, all the faces of a ciihe are s(iHares. 

602. Tlio unit of volume is a eube whose edffe is equal 
to some linear unit, as a cubic inch, a, cubic foot, etc. 

603. The volume of a solid is the uuraber of units of 
volume which the solid contains. 

Being a iiwHiber, a volume may often be dptermitied from other 
numbers in certain expeditious wiiys, wliioii it is one of tlie objects of 
geometry to determine. 

604. Equivalent solids are solids whose volumes are 
equal. 



it number of faces whiol 



El. 2. A square right prism is what JEind of a parallelopiped f 
Ex. 8. Are there more ri^ht parallelepipeds or rectangular paz'al- 

lelopipedst That is, which of tliese indiides tUe other as a special 

<;aser 

Ex. 4. Prove that it a given straight line is perpeinlicular to a 
Ten plane, and auotber straiffht linn is perpendicular to another 
i-aae, aad the two planes are parallel, Iheu the two given lines ara 



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BOOK VII. SOLID GEOilETHY 



Proposition I. Tiieoiiem 

605. Serliona of a piiam vtadc hij pantUd planes cuUing 
all the lateral edges are equal pohjijvns. 




Given tlie prism PQ out by || planes forming the sections 
AB and A'B'. 

To prove section ^7J = secttion A'B'.- 

Proof. AB, BC, CI), etc., arc || A'B', B'C, G'ly, etc., 
respectively. Art. 53i. 

.-. AB, BC, CI), etc., are equal to A'B', B'C, C'D'^ 
etc., respectively. Art.isr. 

Also A ABC, BCD, etc., nre eqiuil to AA'B'C, B'L'l)', 

etc., respectively. Art. 538. 

.-. ABCDE^A'B'C'D'E', Art. 47. 

Q. E. B. 

606. Cor. 1. livery section of a prism made by a plane 
parallel to Ike base is equal to the base. 

607. Con. 2. All fiylU sneliuns of a pt-ism are equal. 



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PRISMS 365 

Proposition II. Theorem 

608. The lateral area of a prism is equal to the product 
of the perimeter of a right section hy a lateral edge. 




BX^aCJ GJ^BCXE, 
area £17 IQ = CT>XE, etc. 



Given the prism KQ, with its lateral area iloiioteil by 8 
and lateral edge by E; and AD a right section of the givou 
prism with its perimeter denoted by P. 

To prove S^PXE. 

Proof. In the prism liQ, each lateral edge = /7, AiLTjOI, i. 

Also AB ± GH, BG X IJ, etc. Art. 505. 

Hence a.vez.CJ RE^AB X GE^AB X ^,1 

■ni^' ^ V Art. ?.Sr>. 

But 8, the lateral area of the prism, equals thi> sum of 

the areas of the ZX7 forming the lateral surface. 

.-. adding, 8'^UB + BG+ CV + etc.) X E. Ax. 2. 

Or S^PXE. 

q. E. D. 

609. Cor, The lateral area of a right prism eQv,als the 
product of the perimeter of the b<tse hy the altitude. 

Ex, Find the lateral area of a rit'lit prtam whose nltitiide U 12 in., 
and whose buso is an equilateral triangle witti a side of G in, Alsa 
flad the total area of this lijfuro. 



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otlfl BOOK Vn. SOLID GEO^IETItY 

PiiorosiTiox III. Theorem 

610. If in-o pnf.ms Jiave the lliirp fa-'cs mcludln// a 
trihi'iJrid angle of one eqmd, rf^pectireli/, to the three faces 
iiidiidhifi a trihedral angle of the oOier, and similarly 
placed, the prisms are equal. 




Given tlie prisms AJ and A'J', having the faces AS, 
AJ>, AG equal to the faces A'K', A'D', A'G', respectively, 
suA similai'ly placed. 

To prove AJ=A'J'. 

Proof. The face d EAF, EAB and BAF arts equal, 

respectively, to the face d E'A'F', E'A'B' and B'A'I". Hyp. 

.-. trihedral ZA = trihedral AA' Art. 584. 

Apply the prism A'J" to the prism AJ, makine each of 

the faces of the trihedral /.A' coincide with corresponding 

equal face of the trihedral LA. Geom.Ai.2. 

.". the plane F'J' ivill coincide in position witJi the 

plane FJ. Art. 500. 

{/"I- tliepnbils C, P, K' coincide Kith G, F, K, respectively). 
Also the point C will coincide with the point C. 

,', Cfl'wili take the direction of CH. Geom.Ax. 3. 
.'. S' will coincide with H. Art. 508, i. 

In like manner J' will coincide with J. 
Hence the prisms AJ and A'J' coincide in all points. 

.-. AJ=A'J'. Art, 4T. 

611. Cor. 1. Two truncated prisms are equal if the 
ihrep. faces i)icliiding a trihedral angle of one are equal to 
the thr&e faces including a trihedral angle of the other. 



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612. COK. 2. Two right prmns are equal if they have 
egiMii bases and equal altitudes. 

Proposition IV. Theorem 

613. An oblique priavi is equivalent to a riqht prism 
whose base is a right section of the oblique prism and whose 
altitude is equal to a lateral edge of the oblique prism. 




Given the 'oblique pvism AD', with the right section JV"; 
also the rigrht priKin FJ" whose lateral edges are each equal 
to a latei-al edge of Alf. 

To prove AD' ^FJ'. 

Proof. AA' = FF. Hyp. 

Subtracting FA' from each of these, AF=A'F'. ("Why ?) 

Similarly BG = B'(}'. 

Also An = A'B', and FO^FO'. Art. 155. 

And A of fare iH7 = homologous A of face A'G'. Art, 130. 
.-. face AW^f.iee A'G', Art 47. 

(fur ihey haco all tlieir parts eijuul, each to each, and .'. can he made 

In like manner face -iK^=face A'K'. 
Eat faee Ai) = Eace A'ly. Art. 5B1, 3. 

.■. truncated priam j4J=tninc.ated i)rLsm A'J'. Art, oil. 
To each of these equals add the solid i-7/. 

.-. AB'^FJ'. (Why?) 

Q.B. ». 



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3G8 EOOK VII. SOLID GEOJIETUY 

PfiOTOSITlON V, ThEOEEJI 

614. The oppo^Ht Uiferal facfs of a paralli'Iop-ipcd are 
equal U)id parallel. 



Given the piiralleSo piped AH with tiie base AG. 
To prove AG = an.d \] EH, anil AJ= aud || BH. 
Proof. The base AC is a /ZJ . Art. 598. 

.-. An = am\ !! EC. (Wiiy?) 

Also the lateral face AJ is a CH . Art. 591, 2. 

.-. ^F^and || EJ. (Whjf) 

.-. inAF=ZCEJ. Art. 539. 

Aua £Z7AG^/-y EH. Art. m. 

Also pUne A(7 11 plane EH. Art. 538. 

In like manner it inay he shoivii thitt AJ and HH are 

eqnal and parallel. 

0. E. D. 

615. Con. All// tiro oppoi^ite faces of a pandlelopiped 
may be lalrn «n the ba.'ics. 

Ex. I. How roBny edKes has a paruHelopiped ! How many faoea ? 
How many dihedral angles ? How many trihedral angles f 

Ex. 2. Find the lateral area of a prism whose lateral edge is 10 

Ex. 3. Find the lateral area of a right prisni whose lateral edge >a 
IG and whose base is a rhombus with diagonals of 6 and 8 in. 



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Proposition VI. Theorem 

616. A plrnie passed throngli liro didgiyii'dhj opposite 
edges of a paralklopiped divides the parallelopiped into two 
equivalent triangular prisms. 




Given the parallelopiped AE with a plane passed through 
the diagooally opposite edges AF and CS, forming the tri- 
angular prisms ABO-G and ABO-E. 

To prove ABC-G^^ADC-K. 

Proof. Construct a plane X to one of the edges of the 
prism forming the right section PQRS, having the diagonal 
PH formed hy the intersection of the plane FHCA. 

Then PQ 11 Sn, and QR \\ PH. (WhyT) 

.-. PQRS is a ZZ7 . (Why ?} 

.-. A PQB= A PSE. (Why ?) 

But (lie triangular prism J.BC-0 ~ a prism wlinse hase 

is the right section PQR and wliose altitude is AP. Art.6i3. 

Also the triangular prism ADG-K o a prism whose base 

's the riglit section PSIi and whose altitude is AF. (Whyt) 

But the prisms having the eqnal bases, PQE and P8R, 

SQd the same altitude, AF, are eqnal. Art, G13. 

.-. ABC'G ^ AD(J-K. A3 1. 

q. E. ». 



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dii) BOOK VII. SOLID Gr.o:^n:TiiY 

Proposition VII, ThEOI!EJI 

617. If two rectangular parallelopipitiWImvii equal bases 
thru "'■^ ^0 each other as their allihules. 




Given the rcctaiiguhir parallnlo|Hpi.';lB P' and P having 
equal bases and tlie dtitutles A' li' ami AH. 

To prove P' -. P^A'B' -. AD. 

Case I. When ike altitudes AT/ iniil AP, 'are com- 
mensurable. 

Proof. Find 'a eoimnon measure of A'li' and AB, aa 
AK, and let it lie contained in A'Ji' n tiinos and in A}i m 
times. 

Then A'B'_: AB^n -. vi. 

Tlii'ough the points of division of A'B' and AP> pa>,3 
planes parallel to the bases. 

These planes will divide I" into ii, and P into m small 
reetansjukr parallelopipeds, all erpml. Ai-t. 612. 

:. r : P = n: m. 
.: F : P==A'B' : AB. {Why ^ 

Case II. When the altitudes A'B' and AB are incoin- 
mcn.'iurable . 

Lut the pupil supply the proof, using the method of 
limits. (See Art. 554). 

618, Def. The dimensions of a rectangular parallele- 
piped are the three edges whieh meet at one vertex. 



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l'AKAI.LELOFIPEDS 



619. Cor. If two rectangular partiUelopipeds haw two 
dimensions hi common, they are to each other as their third 
dimensions. 



PHOPOSITION VIII. TlIEORRM 

620. Two rectangular ■jxirallclo'pipcds having equal alti- 
tudes are to each other as their bases. 




Given the rectangular parallelopipeds P and P' having 
the common altitude a, and the dimensions of their hases 
6, c and &', c', respectively. 

P ^ hXe 
P' b'X c' 

Proof. C«iii^tri.R't 1 he rectangular parallulopiped, <i>, whose 
altitude isit and the dimensions of who^e base are h and c'. 



To prove 



Then 



Moltiplj'ing the corresponding mciubers of thes 
equalities, 



621. Cob. Two rectangulay paralldopipedH having one 
'U'ltiension in common un- to each other as the products of the 
other iiro diiiicnsioiis. 



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iSiJ BOOK Vn. SOLH) liEOMIiTKY 

Proposition IX. Theorem 

622. Ai"j Ih-o rpciangulnr pfirnllelopipeds are to each 
other as tlw pro'liic/a of ilinr thne dhi 



Given tht! rectaiigulai- parailuloplpeds Paiid 2" havir 
the dimensions «, h, c and a', V, e', respectively. 



To prove 



' X fi' X >:' 



Proof. Coiistnict the reetaugular panillolo piped Q hav- 
ing the dimensions a, li, c'. 

P r. 
Then 7}^~' Art. cis. 

Multiplying the corresponding members of these 
equalities, 

P_ » X&X c 

P a'Xb'Xc'' ^''*' 

Q- E. ». 

Ex. 1 . Find the ratio of the volumes of two reetaagular paraOelo- 
pipeds whose edges are 5, G, 7 in. and 7, 8, 9 in, 

Ex. 2. Which will hold mote, a bin 10 s li 1 7 ft., or one Sx 
4 s 5 ft. 1 

Ex. 3. How miiuy Liicks 8x4x3 in. are neeesaary to build a wall 
80s6 ft. sH ja. ! 



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PAliALLELOPirEDS 



Proposition X. Theorem 



623. The volume of a rectangular paralleJopiped is eqtial 
the product of itn three dimensions. 



Given the rectangular para lie loin [jed F having the three 
dimensions a, 6, c. 

To prove volume of P~a X 6 X c. 

Proof, Take as the unit of volame the cube C, whose 
edge is a linear unit. 

Then - = aXbXc ^ 

U 1X1X1 

The volnoie of P is the number of times P containK the 
unit of volume U, or —  Art. 603. 

.". volume of P=« X 6 X i;. 



624. Cor. 1. The volume of a cube is the ciihe of its 
edge. 

625. Cor. 2. TM vobime of a rectangular paraUelo- 
piped is equal to the product of ita base by its altitude. 



Ex, 1. Find the numljer of cubie iuchtg in the volume of a cube 
whose edge is 1 ft. 3 iu. How many bushels dots this box contain, it 
1 bttshel=2150,42ou. in. t 

Ex. 2. Tlie meaBurempiit ot tiie volume if a OAibe reduces to the 

it of the length of what aiiigiu striiiglit line f 



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374 B0(1K VII . SOLIll GEOMETRY 

PROPOSITION XI. THBOai:.1I 

626. The volume of any pamllelopiped is' equal to th^ 

product of its base by its altitude. 




Given the oblique pai-al!t?lopiped P, with its base denoted 
by B, and its altitude by H. 

To prove volmne of P=7J X E. 

Proof, In P produce the edge CD and all the edges 
parallel to CD. 

Ou CJ) produced take FQ^CB. 

Pass planes through F and G ± the produced edges, 
forming the parallelepiped Q, wltli the rectangular base 
denoted by B'. 

Similariy produce the edge GI and all the edges !1 GL. 

Take IK=OI, and pass planes through I and A' X the 
edges last prodneed, forming the rectangular parallclopiped 
B, with its base denoted by B". 



Tlion 




P. 


>e^ 


>R. 






Art, G13 


Also 




£< 


sP' 


-P". 






Art. 3S6 


But 


volume 


ofB 


.B 


XP. 






Art. 625 




.'. volume 


ofP- 


.B' 


Xff. 






Ax. 1 


Or volume 
[OutliEe Proof. F 


OtP 
=.« = 


-P X U. 
S^B" KM. 


= P 


Xil.] 


Ax. 8 
«. B. P. 



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Proposition XII. Theorem 

627. The volume of a triangnlur prism is equal to the 
product of Us hu^e by its aUitude, 




Given the triangalar prism PQB-M, witli its volume 
denoted by 1', area of base by B, and altitude by JI. 

To prove Y=BX II. 

PQ, Qii. Q^J^, construct the 



Proof. ITpou the edg 
parallelopiped QK. 



Hence (^ff^twice FQH-M. 

But volume of QK=in-iin PQBTX R. 
= 2B X H. 
.: twice volume PQR-M==2B X H. 
:. volume PQRSI^ B X H. 



Ex. 2 Find the volumo o! a 
and tLo edata ol whose base ar 



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376 HOOK VII. SOT.TD GK(DMETHY 

PROPOSITION XIII. Theorem 

628. TJie fnltinie of (inij prism is vqual to the produU of 
Us base bi/ its altitude. 




Given tlie prism AK, with its vohime dmioted by V, 
area of base by B, and altitude by H. 

To prove V=BXE. 

Proof. Through any lateral edge, as AF, and the dlag- 
onaJs of the base, AG and ylJ), drawn from its foot, pass 
planes. 

These planes will divide the prism iuto triangular 
prisms. 

Then V, the volume of the prism ,-17i, equals the sum 
of the volumes of the triaugnlar prisms. As. G. 

But the volume of each triangular prisni = ita base X 11. 

Hence the sum of the vohiiues of the A prisms = the 
sum of the bases of the A prisms X II. 

= liy.U. Ax. 8. 

.-. V=BXE. kn. 1. 

<}. E, D, 

629. Cor. 1. Two prisms are to each other as the pro- 
ducts of their bases by their altitudes ; prisms having equiva- 
lent bases and equal altitudes are equivalent. 

630. Cob. 2. Prisms having equivalent bases are to 
each other as their altitudes; prisms having equal altitudes 
are to each other as their bases. 



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377 




PYRAMIDS 

681. A pyramid is a poiyhcdrou 
bounded by a. group o£ planes passing 
through a common point, and by another 
plane cutting all the planes of the group. 

632. The base of a pyramid is the 
face formed by the cutting plane; the ' 
lateral faces are t!ic faces formed by the 
group of planes passing through a com- 
mon point; the vertex in the common point through wliii-h 
the group of planes passes; the lateral edges are the inter- 
sections of the lateral faces. 

The altitude of a pyramid is the perpeodicular from the 
vertex to the plane of the base. 

The lateral area is the sum of the areas of the lateral 
faces. 

638. Properties of pyramids inferred immediately. 

1. The hdn-al fan:^ of a pi/nuiiid arc Iriau'jU'K {Art. 
508, 2) . 

2, Tlie haufi of a pyramid is apnlijgoii (Art. 50S, 2). 

634. A triangular pyramid is a pyramid whose base is 
a triangle; a quad angular py am'd 'a '1 lose 
base is a quadrila 

A triangular pyram a d b U 

faces. All these fa a ay miuya 

taken aa tha basf. 

635. A regu ar pyram d 
is a pyramid who b 
regular polygon, ud fo 
of whose altitu 
with the wiiLcr cf h b 




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378 BOOK VII. SOIJD GEOJIirSRY 

636. properties of a regular pyramid inferred immedi- 
ately. 

1. T]te hiieral ei^yc.'i of a rajidur pymmkl are equal, for 
they are oblique lines drawn from a point to a plam; 
cutting off eqnal distances from tlie foot of the per- 
pendicular from the point to the plane (Art. 518) . 

2. The lateral fwes of a regular pi/ranud are cqiiaj igifs- 
celes Mangles. 

637. The slant height of a regular pyramid is the ahl- 
tude of any one of its lateral faci's. 

The axis of a regnlar pyramid is-its aititode. 

638. A truncated pyramid is thu portion of a pyramid 
included between the base and a section cutting all the 
lateral edges. 

639. A frustum of a pyramid is the /fMT^ 
part of a pyramid included between the //M^-lw 
base and a place parallel to the base. \gl.-...>J"' 

The altitude of a frustum of a pyramid is 
the perpendicular distance between the planes of its bases. 

640. Properties of a frustum of a pyramid inferred Im- 
mediately. 

1. the lahral faces uf a frasiam of a pyramid are 
ira peloids. 

2. The lateral facet! of a frnglnm of a rerjulur punnnid 
are equal isosceles trapesoids. 

641. The slant height of the frustum of a regular pyra- 
mid is the altitude of one of its lateral faces. 

Ex. 1. Sliow that tlie foot of tlie nltitude of a regular pyramid 
eoineides with the center of the cirele eircumscribed abont the base. 

Ex. 2. The perimeter of the midaeetlon of the [rustum of a pyra- 
mid equals oue-iuilf tha aura of the potimutm'a of tlie bases, 



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rvRAMiDS 379 

Pboposition XIV. Theorem 

642. The lateral area of a regttlar pyramid is equal io 
half the product of (he slant height hij the perimeter of the 




Given 0~ABCI)F a regukr pyi'amkl with its lateral 
area denoted by S, slant heiglit by L, and perimeter of its 
base by P. 

To prove g=J£XP. 

Proof. The lateral faces OAJi, O/iO, etc., are equal 
isosceles A. Art. GoG, 3. 

Hence each lateral face has the same slant height, L. 
:. the area of each lateral face = -} L X its base. 
.-. the sum of all the lateral faees = i L X sum of bases. 
^hLX P. 
:. *S' = i LX P. A^,8, 



643. CoK. 77te Intend urea of 
thefrustnm of a regular puraviid is 
equal io one-half the mm of the 
perivieters of its iases multiplied 
by its slant height. 




Ex. Find tlie lttterii.1 ii 
lieifj'lit is 32, uiid an l-S 
uvea liXici, 



I of whosi; base ia lli. Piad tho 1 



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]iOOK VII. SO"LtI) OEOJIETltY 



Pkopositiom XV. Theokem 
644. If a pyramid is cut by a j)?D«e paruUi'l to the fi«se, 

I, The lateral edges and the altitude are divided pro- 
poriionalbj; 

II. Tke section is a polygon similar to the base. 




Given the pyramid S-ABCDF, with the altitude SO cut 
by a plane MN, which is parallel to the base and intersects 
the lateral edges in a, b, c, d, f and the altitude in o. 
So 



To prove I. — =^ 



11. 



8 A SB SC 

The section 
ABCDF. 



SO 
abcdf similar to the base 



Proof, I. Pass a plane through the vertes S II MN. 
Then SA, SB, SC . . . SO, are lines intersected by three 
I planes. 

Sa _Sb _Sc _ ___ S 



 SA SB SO 



SO 



Art. 539. 



II. 



{Wiiy f ) 



ab II AB. 
.: A Sab aud SAB are similar. 
In like manner the A Sbc, Scd, etc., are similar to the 
SBC, SCD, etc., respectively. 

\sb)^ 



 AB 



~BC Ksc) 



yGoosle 



oSrf/ 
ABCDF 




SA^SO " 


«6" 


cihcif . 


. &>' 



PYRAMIDS 381 

That is, the homologous sides of abcdf and ABCDF ave 
proportional. 

Also ^afjc = Z ABC, Ihal = I BCD, etc. Art. 538. 
.•. section aferf/ is similar to tlio buse ABCDF. Art. 32!. 

Q. E. D. 

645. OoE. 1. A section of a pyramid parallel to the 
hose is to the base as the square of its distance from the 
vertex is to the square of the altitude of the pyramid. 



But Ji:^^^ = 



ABGDF SO- 

646. CoE. 2. If two pyramids having equal altitudes 
are cut hy a plane parallel to their bases at equal distances 
from the vertices, the sections have the same ratio as the 
bases. 

Let S-ABCDF and V-PQE be two pyramids cut as 
described. 

n,, abcdf Ho' , , , par Yi 

Then ^— =;=r- Art. G45 ; also -i-i— = ^^- WhjT 

ABCDF «o- DQR VT 

But rr=SO, and n = So. (Why?) 

. »&«?/ ^ W ohrdf_ABCDF 

•■ABCDF PQli °^ pqr PQR ^ " ' 

647. Cor. 3. If two 2>yrHmids have equal altitudes and 
'equivalent bases, seetions iiiiide hy phnies parallel to the bases 
at equal distances f rum Ihv verlk-vs art equivuhnt. 



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KOLID (iEOMETRY 



PhOI'OSITION XVI. TlIKOREM 

648 Tlifi lohaiic of k tn mgitlai j^yrmmd is the limit of 
i Hin "I the loliiwcs of n sciipf of inscribed, or of a series 
f iKtni''' hIkcI ;;iisws of qtml altitude, if the number of 
hi. tndif.miibj inueat,€i^ 




Given the triaogiilar prism 0-ABG with a series of in- 
scriijed, and also a series of circnmseribed prisms, formed by 
passing planes which divide the altitude iuto equal paris, aud 
by making the sections so formed first upper bases, then 
lower bases, of prisms limited by the next parallel plane. 

To prove O-ABG the limit of the sum of each series, if 
the number of prisms in each be indefinitely increased. 

Proof, Each inseribad prism equals the circumscribed 
prism immediately above it. Art. 629, 

.■. {sum of circnmseribed prisms) — (sum of inscribed 
prisms) = lowest civcumscribed prism, or ABC-K. 

If the number of prisms be indefinitely increased, the 
altitiide of each approaches zero as a limit. 

Hence volume ABG-K=0, Art. 253, 3. 

{for Us base, ABC, is cojistant Khile Us altitMe = ) . 
.■, (snm of circumscribed prisms) — (sum of inscribeil 
prisms) = 0. 

.■. volume 0~ABC — (either series of prisms) ^ 0. 
{for this difference < (ajyereiice between the two series, vihick 
lust difference = 0). 
,■, 0-ABC is the limit of the sum of the volumes ot 
either series of prisms. q. e. b. 



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Proposition XVII. Theorem 

649. If two triangular pyramids have equal altitudes 
and equivalent bases, tJiey arc equivalent. 




Given the triangular pyramids 0-AEG and 0'~A'B'C' 
having equivSreQt bases ABC and A'Ji'C, and equal 
altitudes. 

To prove O-ABC'o.O'-A'B'C. 

Proof. Place the pyramids so that they have the com- 
inou altitude R, and divide H into any convenient nnm- 
ber of equal parts. 

Through the points of division and parallel to the plane of 
the bases of the pyramids, pass planes cutting the pyramids. 

Using the sections so formed as upper bases, inscribe a 
series of prisms in each pyramid, and denote the volumes 
of the two series of prisms by V and I". 

The sections formed by each plane, as KLM and K'l/M', 
are equivalent. Art. 6*7. 

.'. each prism in 0-ABC o corresponding prism in 
O'-A'B'C (Art, C29). .-. F= F. As:. 2. 

Let the number of parts into which the altitude is 
divided be increased indefinitely. 

Then V and V become variables with (h-ABC and 
0'~A'B'C' aa their respective limits. Art. MM. 

But I'oT'' always. (WLv?) 

:, 0-ABC~0'-A'B'C'. (Why!) 

(1. X. D. 



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BOOK V)I. SOLID 



PHOl'DSITrON XVIII. TlIEORF,:« 



650. Thr vohinK' of a fridDditlar pyramid i 

oiif-lhird Ihr iirodiid of iV.v bam: hy il^ nUilmh'. 




Given the triangiilar pyramid 0-AB€, having its volnmo 
denoted bj- V, the area of its base by B, and its altitude by H. 

To prove V^^BXM. 

Proof. Ou ABC as a base, with OB as a lateral edge, 
construct the prism ABG-BOF. 

Then this prism will be composed of the original pjTa- 
mid 0-ABG and the quadrangular pyramid 0-ADFO. 

Through the edges OJ) and OC pass a plane intersecting 
the face ^DFC in the line B<7, and dividing the quadrangular 
pyramid into the triangnlar pyramids 0~ADC and OSFC. 

Then 0-AT)C^O-I>FC. Art. 649. 

{for IkeijMrc the common rerlci 0,<i,nd the equal bases ABC andBFC). 

But 0-DFC may be regarded as having as its vertex 

and DOF as its base. Art. 634. 

.-. (}'DFC=:^0-ABC. Art. 619. 

.'. the prism is made up of three equivalent pyramids. 
.-. (}-ABC=h the prism. As. 5. 

But volume of prism = B X H. Art. 627, 
.-. 0-ABC, or r=iBX H. Ax. 5. 
g. E. B. 

Ek. FJ^id the volume of a trmngula,r pyramid whosa altitude 13 12 
ft,, ami whose b:iao is an equilateral triangle with a bide of 15 ft. 



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Proposition XIX. Theorem 

851. The volume of any pyramid is equal to one-third 
iAe product of itn base by Us ultitude. 




Given the pyraraid 0- ■il>( VF ln\ mg its volumfi denoted 
^>y V, the area of its baae bj £., and its altitude by H. 
To prove F= i B X E. 

Proof. Through any lateral edge, as OD, and the diago- 
nals of the base drawn from its foot, as AS and BD, pass 
planes dividing the pyramid into triangular pyramids. 

Then Y, the volume of the pyramid 0-ABCDF, will 
Bqnal the sura of the volumes of the triangular pyramids. 
But the volume of each A pyramid = J its base X E. 

Art. 650. 

Hence the sum of the volumes of A pyramids=i sum of 

their bases X E. Ax. 2. 

= k I) X E. Ax. 8. 

.-. r=h BX S. Ax. 1. 

Q. E. D. 

852. OoR. 1. 5'fie volumes of two pyramids are to each 

'>iher as the products of tkeir banes atid altitudes; pyramids 

fiaving equivalent bases and equal altitudes are equivalent. 

658. Cob. 2. Pyramids having equivalent bases are to 
^eh other as their altitudes; pyramids having equal alti- 
'Mifes are to each other as their bases. 

854. Scholium, The volume of any polyhedron maybe 
J'^und by dioidiuff the polyhedrmi Into pyramid.'^, finding the 
^hnie of each pyramid •jcpundchj, and tailing their sum. 



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BOOK VII. aOI.ID GEOMETRY 



Proposition' XX, Theorem 

656. The frustum of a triangular pyramid is eguivaleKt 
to the sum of three pyramids whose common altitude is the 
altitude of IM frustum, and tvJioae bases are the lower base 
the upper base, and a mean proportional between the two 
bases of the frustum. 




Given ABG-DEF the frustum of a triangular pyramid, 
having the area of its lower base denoted by B, the area 
of its upper base by b, and its altitude by B. 

To prove ABG-DEF ^ three pyramids whose bases are 
B, h and V Bb, and whose common altitude is H. 

Proof. Through E and AG, E and DC, pass planes divid- 
ing the irustum into three triangular pyramids. Then 

1. E-ABC has the base B and the altitude H. 

2. E-DFG, that is, G-DEF, has the base b and the 
altitude if. 

3. It remains to show that E-ADC is equivalent to a 
pyramid having an altitude H, and a base that is a mean 
proportional between B and 6. 

Denoting the three pyramids by I, II, III, 

I_ A ABE __AB^_AC^ ^ ADC ^11 

II A AI)t:~DE BE A I>FG III' 

(Arts, 653, 391, 644, 321. Let the pupil supply the reaaon for each 



I II 



step in detail). 
(Ax. l.)orII = l/lXiny 



.-. E~ADG=y'{^HXB) (hSX b}=iHVBXb. 
Hence, J.£OZ'iJ#— sum of three pyramids, as deseribecU 
^. £■ S. 



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656. Formula for volume of frustum of a triangular 
pyramid. F=i JI {B+h + VmY. 

Proposition XST. Theorem 

657. The volume of Oie fi-usium of nny pyramid is 
eguivaJent to the sum of the volumes of three pyramids, whose 
common attitude 4s the altitude of the frifsium, and whose 
lases are the lower base, the upper base, and a mean propor' 
tional between the two bases of the frustum. 




Given the frustum of a pj-ramid Ad, having its volume 
denoted by V, the area of its lower liase by B, of Its upper 
^>ase by 6, aud its altitude by H. 

To prove T=h E (B + b + VBb). . 

Proof. Produce the lateral faces of Ad to meet in K. 

Also construct a triangular pyraiuid with base PQR 
■qnivalent to ABCDF, and in the same plane with it, and 
■Tith an altitude equal to the altitude of K-rABCDF. Pro- 
luce the plan€ of. ad to cut. the second pyramid in pqr.  

Then pqr'cahcdf. Art. 647^ 

.-. pyramid ST-ABCPFo pyramid T-PQIi. Art. (552. 

Also pyramid K-abtdf = pyramid T-fqr. (Why ?) 

Sttbtractiug, frustum Ad ^ frustum Tr. Ax. 3. 

But volume Pr^-]. 11 Hi Jrh+VlThh 

:. volume Ad^;, II {n ]-b^VBb).' (Why?) 
-"Q. EiB. - 



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SOLID GEOMETRY 



Piioposrnox XXIL Tiieorkm 

658. A truncuted triangular priim is vqvivalent to tTie 
sitvi of three pumHiids, of which the base of the prism is the 
common base, and whose vertices are the three vertices of the 
inclined section. 





Given the truncated triangiilai- prism ABG- PQR. 

To prove ABC-PQR ^ the sum of the three pyramids 
P-ABC, Q~ABC and Ti-ABC. 

Proof. Pass planes through Q and AC,Q and PC, divid- 
ing the given figure into the three pyramids Q-ABC, 
Q-APG and Q-FBG. 

1. Q-ABG has the required l*ase and the required 
vertex Q. 

2. Q~APC^B-APG, Art.652. 
(/or they liave tkesame base, AFC, and the same altitude, their ver- 

iices beind in a line |i bate APC), 

But B~APG may be regarded as having P for its ver- 
tes, and ABG fov its base, as desired. Art. 634. 

3. Q-PBC<i= B-ARC (see Fig. 2). Art. 652. 
\foT the base ASC ^ base PSC (Art. 300); and the altitudes of the iao 

pyramids are eqnal, the vertices Q and B being in line || plane 
FACB, in lahich the bases lie). 

But B~ARG may be regarded as having R for its ver- 
tex, and ABC for its base, as desired. Art. 63* 



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-•. ABC-PQB o sura of three pyramids whose eommoii 
base is ABG, and whose vertieea are P, Q, R. 



659. Cor. 1. The vobinie of a truncated right triangii- 
iar prism (Fig. -i) '.v '■q/'id to the product of its la^ie hij 
lyne-third the auin of Ur lateral eOges. 

660. Cor. 2. The vohime of any truncated triangular 
prism (Fig. i) is equal to the product of the area of its 
ngM section (ii/ one-third the sum of its lateral edges. 



P8ISMAT0IDS 

661, A prismatold is a polyhedron 
bounded by two polygons in parallel planes, 
called bases, and by lateral faces which are 
either triangles, trapezoids or 



662. A pfismoid is a prismatoid in 
*bich the Itiises have the same number of 
sides and liavt- their corresponding sides 
parallel . 



&"■ The Tolurae of a truncated right parallelopiptd equals the 
•fea ol the lower base multiplied by oue-fourth the sum of tlie lateral 
^i^eea (or by a perpend'ciilar from the ceuter of the upper base to tUe 
lower basul. 




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r,i:inn-ri'i{Y 



Proposition XXIII. TiiKOiiEJi 
The vohmie of a prisituifold is iq'int (o 



the product of its ultiiude hy tin: si 
fowr Hmes the area of its midsection. 



sixth 
of its bases and uf 




Given the prisiuiitoitl ABCI'-FOK, with bases B and h. 
midsection Jf, volume 1', atul altitude Tl. 

To prove Y^\H {B-^h^A 21) . 

Proof. Take any point in tlie midsection, and through 
it and each edge of the prismatoid let planes be passed. 

These planes will divide the figure into parts as follows : 

1. A pyramid with vertex 0, base ABCD and altitude 
i H, and whose volnme .". = ,V 77 X jR. Art. 65] , 

2. A pyramid with vertex 0, base FGK, and altitude i 
if, and whose volume .■. =i 7/ X h. (Why?) 

3. Tetrahedrons like O-A'lKl who^e volume may be 
determined as follows (see Fig. 2): 

AB^-l-FQ. (Why!) 

.-.A .it?B=4 A PGQ. Art. 398. 

.-. O-AGB^i 0-PGQ. Art. 653. 

But 0~PGQ (or G-FQO) = h I'QO X h H = i HXPQO. 

.: O-AGB^i SXiAPQO. (Why?) 

.-. the sum of all tetrahedrons like 0-A(3B=i HXiM. 

:. Y=-\ EXB + i HXh+i 11X4 M. 
Or V=}E iB+h + iM). l!.S.l>. 



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RFGULAE POLYHEDRONS 



REGULAR POLYHEDRONS 

664. Def. a regular polyhedron is a polyhedron all of 
whose faces are equal regular polygons, and all of whose 
polyhedral angles are equal. Thus, the uuhe is a regular 
polyhedron . 

Proposition XXIV. Theorem 

665. Btit five regiihir polyh^dronti are p 




Given regular polygons of 3, 4, 5, etc., s 

To prove that regular polygons of the same number of 
sides ean be joined to form polyhedral <^ of a regular 
polyhedron in but five different ways, and that, conse- 
quently, but five regular polyhedrons are possible. 

Proof. The sum of the face A of any polyhedral angle 
< 360°. Art. 583. 

1. Each Z of an equilateral triangle is 60°. Art. i.ii, 
3 X 60°, 4 X 60° and 5 X 60° are each less than 360°; 

bat any larger multiple of GO° = or > 360°. 

.". but three regular polyhedrons am be formed with 
equilateral &. as faces, 

2. Each Z of a si/Krtc* contains 90°. An. irii. 
3 X 90° is less than 300°, but any larger multiple of 

90" = or > 360". 

.*. but one regular polyhedron enn be formed with 
Squares as faces. 

3. Each i^ of a regular jiewiai/oM is 108°. Art. \--i. 
3 X 108" is less than 300°, but any larger multiple of 

108° > 300°. 



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Jii)^ HOOK VII.' SOLID riEo:Mi:'niv 

.■. but one regular piil.\ lu'dmn u;ui lu' formed wHli regu. 
lar peutagons as faeos, 

4. Eaob Z of a regular iu'xaffoii is 120°, anil 3 X 120" 

.". no regular polyhedron eau be formed with hexagons 

or with polygoiiB with a greater number of sides as faces, 

.'. but five regular polyhedrons are 



666. T!ie construction of the regular polyhedrons, by ths 
nse of cardboard, may be effected as folltjws; 

Draw on a piece of cardboard the diagrams given below. 
Out the cardboard half through at the dotted lines and eu- 
tirely through at the full lines. Bring the free edgea 
together and keep them in their respective positions by 
some means, snch :is pasHnft- strips of paper over tViem. 




TTfM 



P^^ W'i^p -z^ 



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POLYHEDRONS ii))iS 

POLYHEDRONS IK SEHESAL 
PliOPOf^ITION XXY. TllEOKEM 

667. III. inn/ polnhedron , the nitnibi'r of edges increased 
by two eqiuils the number of vertices increused hy the miinher 
offacen. 




Given the polyhedron AT, with the number of its ver- 
tices, edfjes and faces denoted by V, E and F, respectively. 

Toprove E + 2=V+F. 

Proof. Taking the shigle face ABCD, the nuraber of 
edges equals the number of vertices, or E= V. 

If another face, CRTT)., be annexed (Fig, 2), three new 
edges, 6'i?, BT, TI), are added and two new vertices, B and T. 

:. the mimber of edges gains oue on the aumber of ver- 
tices, or E^ T+ 1. 

If still another face, BQRC, beflunexed, two new edges, 
BQand Qli, are added, and one new vertex, Q. .: E= V+2. 

With each new face that is annexed, the number of 
edges gains one on the number of vertices, till but one 
face is lacking. 

The last face increases neither the number of edges nor 
of vertices. 

Hence number of edges gains one ou number of vertices, 
for every face except two, the first and the last, or gains 
F~1 m all. 

.■. tor thi- entire tiK'n-c, /■:= I'-f- F—'l. 

That is £.' + 12= t'-i- t\ A.V. 3. 

Q. s. n. 



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.i'Ji ]!OOK vn . SOLID GEOJIETEY 

Proposition XXVI. Theorem 

668. The sum of Ike fncc amjlex of any j'ohjhedron 
equals four right ant/Us iaknn us inunij times, less two, (o,- 
the polyhedron has vtiikfs. 




Given any polyhedron, with the sura of its face angles 
denoted by S, and the number of its vertices, edges and 
faces denoted by V, E, F, respectively. 

To prove 8= { V—2) 4 rt. A . 

Proof. Each edge of the polyhedron is the iatersection 
of two faces, ,■. the number of sides of the faceE = 2 E. 

:. the sum of the interior and exterior A of the faces = 
2 S X 2 rt. A, or EX i rt. A . Avt, 73. 

But the sura of the exterior A of each faee = 4 rt, A . 



:. the sum of exterior A of the Ffaces = i'X4 rt. A. 


Subtracting the sum of the exterior A 


from the sum" of 


all the A, 


the 


Bujn of the interior A • 


of the P faces = 


(.SX4rt. 


A)- 


-{FX4 rt. A). 




Or 




li={E-F) 4rt. A 




But 




E + 2=r+F. 


Art. 6QT. 


Hence 




E~F^V~2. 


As. 3. 


Substituting 


■,tor E—F, S=(V~~2) 4 r 


t. A . Ax. 8. 








Q. I. B. 


E5, Veri 


f J the laal two tlieorema iu the uase 


of the cube. 



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COMPARISON OF POLYHEDKONS 



COMPARISON OF POLYHEDKONS. SIMILAR POLYEEDKOWS 

Proposition XXVII. Theoreji 
669. // two tetrahedrons have a trihedral angle of one 
equal to a trihedral angle of the otJter, they are to each other 
as the products of the edges including the equal trihedral 
angles. 




Given the tetrahedrons O-ABC aud O'-A'B'C, with 
theirvohiines denoted by Faad V, respectively, and having 
the trihedral A and 0' equal. 

V OAXOBXOC 

To prove yr ^yA' X O'B' X O'C' 

Proof. Apply the tetrahedron O'-A'B'C to (h-ABC so 
that the trihedral Z (y shall coincide with its equal, the 
trihedral Z 0. 

Di-aw CP and O'F 1 plane OAB, and draw OP the 
projection of OC in the plane OAB. 

Taking OAB and OA'B' as the bases, and OP and G'P' 
as the altitudes of the pyramids 0-OAJi and C'-OA'B', 
respectively. 

V A OAB X CP A OAB ^ CP 



But 



A OA'B' XC'F A OA'B' 
A OAB OA X OB 



-Xr 



07" 



Art, <ir>'> 



OA' X OB' 

CP _ OC 



A OA'B' 

In the similar rt. A OCPand OC'F, 

 T^ OAX OBX OC ^ OA X OB X OC 
■'■ V" OA' X OB' X OC O'A' X O'B' X O'C' 



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?,nc, 



SOLID GEOMETKV 



670. Dep. Similar polyhedrons are polyhedmus having 
the saaie iinmbei' of (aces, similar, each to each, and simi, 
lai-ly placed, and having thcii- eorrespondiog polyhedral 
augles equal. 

PKOi'OSiTio^- XXVni. ThEORG-M 

671. Any iiro s-iimlar polyliedrons way he (leeompoaed 
into the t:ftme niimle.f of lef raited rons, similar, each to each 
and similarly placed. 




Given P and P', two similar poiyhfidrons. 

To prove tlv;it P and P may be decomposed into the 
same number of tetrahedrons, similar, each to each. 

Proof. Take R and W any two homologous vertices of 
P and P'. Draw homologous diagonals in all the faces of 
P and P except those faces which meet at H and H', sepa- 
rating the faces into corresponding similar triangles. 

Through H and each face diagonal thus formed in P, 
and through S' and each face diagonal in P', pass planes. 

Each corresponding pair of tetrahedrons thus formed 
may be proved similar. 

Thus, in E-ABC and II'-A'B'C, tlie A EP,A and 
H'B'A' are similar. Art. 329. 

In like manner A HBC and H'B'C are similar; and &■ 
ABC and A'B'C are similar. 



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SIMILAR POLYGONS 397 

., NA /HB\ EC /BC\ AC . „. 

.-. A ARC and A'WC .-u-e simiiar. Art, 326. 

Hence the corresponding faces of H-ABC&mlH'-A'B'C 
are similar. 

Also their homologous trihedral li are equal. Art. 5S4. 

.". tetrahedron S-ABC is similar to H'-A'B'C. Art. 670, 

After removing S-ABG from P, and H'-A'B'C from 

P', the remaining polyhedrons are similar, for their faces 

are similar, and the remaining polyhedral A are equal. 

As. 3. 
By continuing this process, P and /" may be decom- 
posed into the same number of tftrahedroos, similar, each 
to each, and similarly placed, 

Q. E. D. 

672. Cor. 1. The hoDwloijoim vdyes of similar polyhe- 
drons are proportional; 

Any two homologous line'i in iiro aimilar polyhedrons 
iiave the same ratio ws any other tiro homologous Hwh. 

678. Cor. 2. Any tiro liiiimloijous faces of iiru .similar 
polyliedrons are to each other «« the nquaret: of any tivo 
homologous edges or lines; 

The total areas of any tiro similar polyhedrons are to 
each other as the sguares of any two homologous edges. 

Es. 1. Iiithefigure.p.39,i,if tl 
and those meeting at 0' ave *, 6, : 
of the tetrahedrons. 

Kz. 2, If the linear dimeasions of one room are twice as great as 
the eorrespondiag dimensions of another room, how will their enrf aces 
(and ,". cost of papering) compare 1 liow will thi'iv volumes compare T 

Ex. 8, How many 2 iu. cubes can be cut from a 10 in, cube t 

Ei. 4, If the hasL's of a prisraoid are raetansl^s whose dimensions 
we 0, b a:id b. a. and altitude is E, And the formula for the volume. 



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674. 

each otii 



BOOii YII. SOLID GEOMETRY 

PROPOSITION XXIX, Thbokeji 

Till', volume:; of tiro similar tdmhcdrons 
r ns the cull's of any pair of Jioniologous 




Given the similar tetrahedrons O-ABC and O'-A'S'G'. 

To prove — — -.m:-,' 

1" O'A'' 

^ . y OAX OBXOG ,^, „, 



V (yA' X O'B' X O'C 

^OA y OB y OG 

O'A' O'B' O'c' 

OA ^ OB __ PC 

O'A' O'B' O'G' 

V OA yO^y^ OA ^ ~0A^ 

' r'~0'A' O'A' O'A' ooT'^ 



El. 1. In the above figures, if AB=^2 A'B', fiud the ratio of, V 
to r. Find the same, if AB^U A'B'. 

Ex. 2 The meaBUremeDt of the volume of a, regular triangular 
prism reduces to the laeasurBment of the lengths of how many straight 
lineB ! of a fmstHin of a regular square pyramid ? 

Ex. 3. Show how to eonstruet out of pasteboard a iBgular pli^m, 
B puallelapiped, aad a truncated siiua.re prUtn. 



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similar polygon's s 

Proposition XXS. Theorem 

675. The volumes of any two similar polyhedrons i 
to each other as the cubes of .any two homologous edges, 
of any other two homologous lines. 




Given the polyhedrons AK and A'K' having their vol- 
umes denoted by Fand V, and RB and B'B' any pair of 
homologous edges. 

To prove — "^ "•-.:".•:' 

Proof. Let the polyhedrons be decomposed into tetra- 
hedrons, similar, each to each, and similarly placed. Art. 671. 

Denote the volumes of the tetrahedrons in Pby t-i, vs, 
i-'i . . . and of those in J" by f'l, v'2, v'-i . . . 



Then 






V': 


1 H'B' 










Also 




(/"• 


each of 


these rat 


''°'~Wb' 


j) 


Art. 


674, As. 1. 


" t 


n + t 


'1 + 


«'. + - 


- ^\' 


that is, 


V 


.^1 


 Art. 3la, 



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iW HOOK Vil. SOLID GLO^TE'l'in: 

TWKOitEMS rdXCEHNJNC POLVHEOROXS 
Ei. 1. TliB Iriteral faees of :i rigbt prism are rectangles. 
Ex. 2. A diagonal plaue of ti prism is parallei to every lateral edge 
of the prism not coiitnined in the plnue. 

Ex. 3. The diagouais of a parallelepiped bisoci^ each other. 

Ei. 4. The square nf a diagonal o£ a reijtiingular parallelepiped' 
oqaaJs the sum of the squares of the thrive edges meeting at a vertex 

Ei. 5, Each lateral face ol :\ prism is pavallel to every lateral edgo 
not contained in the face. 



Ex. 
lateral 


6. Every section of a prism made by a plane parallel to a 
Bilge is a parallelogram. 


Ex. 
allel ti 
prism 


7. It any two diagonal planes o' 
D each other are perpendicular t 
is a right prism. 


the base of the prism, the 


Ex. 

Iflse is 


8. What part of the volume of 
. a face of the cube and whose vei 


3 cube is the pyramid 
■fex is the center of the 


whose 
onbef 


Ex. fl. Any section of a regular squ 
throngh the asis is an isosceles triangle. 


are pyramid made by a 


plane 


Ex. 10. lu any regular tetrahedron, 
the perpendicular from its foot to any !'; 


an altitude equals three 


times 


Ex. 


1 1. Id any regular tetrahedron, 


, ail altitude equals the i 


.urn of 



the perpendiculara to the faces from any point within the tetratedron. 

Ex. 12. Find the simplest formula for the lateral area of a trun- 
cated regular prism of n sides. 

Ex. 13. The sum of the squares of the four diagonals of a paral- 
lelepiped is equal to the sum of the squares of the twelve edges. 
[Sno. Use Art. 352.] 

Ex. 14. A parallelopiped iij symmetrical with respect to what point? 

Ex. 15. A rectanglar parallelo piped is symmetrical with respect to 
how many planes? (Let the pupil make a definition of a figure syni- 
metrical with respect to a plane. See Arts. 4S6, 487.) 



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EXERCISES OX POLYHEDHOXH -JOl 

Ei. 16. The volumfl of a pyr'amid ivbose lateral odges mn the 
three edges of the parallelepiped m«eting at a point is what part of 
the volume ot the parallelopiped 1 

Ex. 17. If a plane be passed through a vertex of a cube and tlie 
tliagonal of a faae oot adjacent to the vertex, what part of the volume 
of the enbe is eoataioed by the pjrainiil so formed ! 

Ex. 18. If the angles at tho vertex of a triangular pyramid are 
right angles and each lateval edge equals a, show that the volume of 



Es. 19. IIoiv large is a dihedral anglo at the base of a regular 
pyramid, if tho apotheia ot the base equals the altitude of tho pyramid 1 



the lateral surface. 

Ex. 21. The section of a triaiiguiar pyrainiil by a plane parallel 
to two opposite edges ia a parallelogram. 

If the pyramid is regular, what kind of a parallelogram does tha 
section become ? 



Ex. 22. The altitude of a regul; 
ot the base into seguients whii.'h ari 


:ir tetrahedron 


divide! 


Ex. 23. If the edge of a reguh 
elant height is ^ ; and hence th 


ir tetrahedron 
at the altitude 


'* 3 ' 


ume is ^^' 







Ex. 24. If the midpoints o£ all the edges of 
B- tetrahedron except two opposite edges be 
joined, a parallelogram is formed, 

Ex. 25. Straight lines joining the midpoints 
'^t the opposite edges of a tetrahedron meet in 
s point and bisect each other. 

Ex. 26, The midpoints ot tl^- ed-es of a reg' 
^e vei'.ic..^ ot a regulur octiihedron. 



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JIOOK Vll, SOLID GEOMKJ'HV 

EXERCISES. CROUP Ca 

intOELKMS CONrKRSING POLYHF.DBONfi 

isect the volume ot a given [insm by a plane parallel U 



Ex. 3. Through a given point pass a plane which shall bisoct the 
volume of a given paralleloplpeil. 

Ex. 4. Given an edge, couatruet a regular tetrahedi'ou. 

Ex. 5. Given an edge, eonstruct a regular octahedron. 

Ex. 6. Pass a plane tiirougb the axis of a regular tetrahedron eo 
that the section shiili bo an iaoaceles triangle. 

Ex. 7. Pass -J, plane through a cube so that the section nhall be a 
regular hexagon. 

Ex. 8. Through three given lines no two o£ which are parallel 
pass planes which shall form a parallelepiped, 

Ex. 9. Prom cardboard eonatiuct a regular square pyramid each 
o£ wbose faces is an equilateral triangle. 



EXERCISES. GROUP 69 

REVIEW F.XEKCISES 





Make a 1 


ist of the pr 


operties of 


Ex. 1. 


Straight li 


nos in space. 


Ex. 9, Eight prisms. 


Ex.2. 


One line a 


nd one plane 


Ex. 10. Parallelopipeds in gen- 


Ex.3. 


Two or n 


lore lines an 


,d ^'■'''■ 


one 


plane. 




Ex, 11. Rectangular parallelo- 


Ex.4. 


Two plane 


B and one lin 


e, pipeds. 


Ex.5, 


Two plane 


sand two line 


g Es, 12, Pyramids in general. 


Ex.6, 


Polyhedrons in general. 


Ex. 13, Regular pyramids. 


Ex, 7. 


Similarpolyhedrons. 


Ex. 14 Frusta of pyramids. 


Ex, 8. 


Prisms in 


general. 


Ex, 15, Truncated prisms. 



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Bdox VIIT 
CYLINDERS AND CONES 

CYLiHDEKS 

876. A cylindrical surface ii5 
a carved surface generated bj a 
straight line which moves so :in 
constantly to touch a given fixed 
curve and constantly be pLirLilL-l 
to a given fixed stniigiit line 

Thus, every stiadow cast by a pomt 
of light at a great distance, ao l>y a 
Btar or the sun, appTOximatPs the 
oylindrioal form, that is, is, Ijouaded Cjimdrnal surface 

by a, oylindrieal Hurfaoe of light. Henee, in all radiations (ns of light, 
heat, magDetism, etc.) from a point at a great distance, we are 
concerned with oylindrieal snrfacea and solids. 

677. The generatrix of a cylindrical surface is the mov- 
ing straight line; tlie directrix is the given curve, as CDE; 
an element of the cvlindrical suffice i'; the moving straight 
line m any one of its positions, as BV 

678. A cylinder lo i solid bounded bj 
^ cylindrical sniface md L\ two piialiel 
planes. 

The bases of a c\lmdei ul its puallel 
plane faces; the lateral surface is the 
cylindrical surface iiiclud d liLt\\eeu the 
parallel planes foiming its bT^f s tlic alti 
tude of a eyliudHi is the distiin,e lictwteu the 1 

The elements of a e^lmhi ait the elements of the 
cylindrical surface boundm': it. 




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iO[ 



JiOOK Yin, SOLID GEfniKTKY 



679. Property of a cylinder inferred immediately. All 
the elements of a cylinder are equal, lor they ;ire parallel lines 
included between parallel plai.u's (Arts. 533, 076), 

Tlie cjiinilers most important in practical lite irt, those determined 
by their stability, the ease with wliich tlipy can bi, iiude from com- 

680. A right cylinder is & cyhndci 
wliose elements are perpendicular to tlie 
bases. 

681. An oblique cylinder isoue-nliui 
elements are oblique to the bases, 

682. A circular cylinder is a cjlmd i 
whose bases are circles. 

683. A cylinder of revolution is a c\liii- 
der generated by the revolutiou of a lect 
angle about one of its sides as an axis. 

Hence, a cylinder of revolutioa is a right 
circular cylinder. 

Some of the properties of tills solid ore derived 
most readily Tiy cousidering it as generated by a re- 
volving rectangle ; and otliers, by regnrdiag it an a 
particular kind of cylinder derived from t!ie grnLial 

684. Similar cylinders of revolution ne f\lmdeib gen- 
erateil by similar ructaiigles revolving ibout homologous 
sides, 

685. A tangent plane to a cyliud \ is t, plane whioh 
contains one element of the cylinder, and whi-'h do^!^ not cut 
tlie cylinder on being produced. 

Es. 1. A plane passing through a taugent to the base of a circu- 
lar cylinder and the element drawn through the point of contact is 
tangent to the cylinder. (For it it is not, etc.) 

Ex. 2. If a plane is tangent to a dicular cylinder, its intersection 
villi tlic lilane of the base is tangent to the base. 




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CYLINDERS 

686 A prism inscribed lu a cylinder is a pr 
lateral edge'^ ue ebmeiits of the cylinder, and wnos 
ni I H m tlur- 1 I es of the cylinder. 



405 
prism whose 





687 A prism circumscribed about a cylinder is a prism 
whose litenl fice-. are tangent to the cylinder, and whoso 
bases are polygons eii-enmscribed about the bases of the 
cylinder. 

688. A section of a cylinder is the figure formed by the 
intersection of tlie cylinder by a plane. 

A right section of a cylinder is a section formed by a 
plane perpendicular to the elements of the cylinder. 

689. Properties of circular cylinders. By Art. 441 the 
area of a circle is the limit of the area of an inscribed or 
circumscribed polygon, and the circumference is the limit 
of the perimeters of these polygons; hence 

1. The volume of a circular cylinder is the limit of the 
volume of an inscribed or circuiAscribed prism. 

2. The lateral area of a circular cylinder is the Umil of 
the lateral area of an inscribed or circumscribed priism. 

Also, 8. By metJiodf loo advanaedfor this book, iC may be pravfd Ih-il 
the jierimeUr ol'u rig/it atttuin is the limit of the perimetm- of a rif/hl ner- 
tvm of an insrvib&l or drcinnMri'ied pri»'n. 



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690. Ei-enj section of a rylUuli-r mcuh hi) a plane pass- 
ing tkrou'jh an ekmcid is « parnllrloarain. 




Given the <>ylindpf AQ cut by a ■plixuc passing through 
the element AB and fonriiiig the section ABQP. 

To prove A BQP a CJ . 

Prooi. API! BQ. Art. 631. 

It reraaiES to prove that PQ is a straight line II AB. 

Through P draw a line in the cutting plane ll AB. 

This line will also lie in the cylindrical surface. Art. 676. 
.■. this line must eoineide with FQ, 
(fnr the fine drawn lies in both the cutting phmc and the cylindrical 
surface, liencc, it iiiiiKt lie their iatvrscction). 

:. FQ is a straight line || AB. 

:. ABQP is a /ZJ . (Why?) 

(J. E. D. 

691. Cor. Every section of a rirjht cylinder made, by a 
plane parsing through an element is a rectangle. 

Ex. 1. A doov siTiii!;iii;; on its hinges generates what kind of a 

Es, 2. Every sectioQ of a paBiillelopipeii maJe by a plane iater- 
secting all its la,teral edges is a imrullciogiaw, 



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CYLINDEBS 

Proposition If. Theorem 
The bases of a cylinder are equal. 



Given the cylinder AQ with tlie bases AVB and CQB. 
To prove base APB = base GQD. 

Proof. Let AC and BD be anj' two fixed elements in 
the snrface of the cylinder AQ. 

Take P, any point except A and B in the perimeter of 
the base, and through it draw the element PQ. 
Draw AB, AP, PB, CD, CQ, QI). 
Then ACand Bi> are = and II. (Why!) 

.-. AD is a ^17 . (Why?) 

Similarly AQ and BQ are 07 . 

.-. AB^CD, AP^CQ, and P.P=T)Q. (Why?) 

.■. AAPB=A CQD. (Why?) 

Apply the base APB to the base CQD so tl.at AM coin- 
cides with CD. Then P will coincide with ^, 
(for A AFB = A CQD). 
But P is any point in the perimeter of the base APB. 
.: every point in the perimeter of the lower base will 
coincide with a corresponding point of the perimeter of 
the upper base. 

.'. the bases wiil coincide and arc equal- Art. 47. 

^, E, B. 



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408 HOOK Vm. fiOT.lD fJT:0:METl!Y 

693. Cob, 1. The sscfioit.'i of a cjlinder madp by two 
paraJlfl phities cutting all the elements are equal. 

For the sections thus formed ai'e the bases of the eylinder- 
iucluded between the cutting planes, 

694. Cor. 2. Attn section of a cyHmhr paniUel to the 
base is equal lo the bat^e. 

pROPOsiTiON III. Theorem 

695. TJie lateral urea of n cirmdnr vylimler is equal fo 
tlie product of the perimeter of a ritjltt section of the cylin- 
der by an elemertf. 




Givea the circular cylinder AJ, haviu" its lateral area 
denoted by 6', an fiknifnt by JS, and the perimeter of a 
right section by P. 

To prove S=PXE. 

Proof. Let a prism with a regular polygon for its base 
be inscribed in the cylinder. 

Denote the lateral area of the inscribed prism by 8', 
and the perimeter of its rijjht section by P'. 

Then the lateral edge of the inscribed prism is an ele- 
ment of the cylinder. Constr 



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4ft. asa, 3. 


(Whyf) 


(Why?) 
Q. E. D. 



CYLINDKKS 409 

.-. S'^PXE. Art. COS. 

If the number of lateral faces of the inscribed prism be 
indefinitely increased, 

iS" will appi'oaeh ^ as a limit. Art. 689, 2. 

P' will approach P as a limit. Art. 689, 3. 

And P'Xfiwill approach PXE as a limit. 
But. ,S" = PX.E always. 

.-. 8 = PXE. 

696. Cor. 1. The laieml area of a cyliwhr of involu- 
tion is equal I" the produci of Ihe civionferpnee of its base, 
by its altHiirh. 

697. Formulas for lateral area and total area of a cylin- 
der of revolution. Denoting the lah-ral area of a cylinder of 
revolution by S, Ike total area by T. the radius by R, and 
the altitude by U. 

T=2 TtRH + -2 nR" :. T^2 nR {H^ R). 



Ex. 2. If the a 
of the Lase (H = l 
terms of li t also, in tenns nf // f 

Ex. S. What do they liecorae, if the altitude eqwals tlie diametsr 
<rt the base ? 

Ex. 4. Ill a cyliniier of revolution, what is the ratio of the latersl 
"^ea to the area, of the base f to the total area ! 



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4H) BOOK VTII. SOLID OLOMKTP.V 

I'HOropmuN i\. TiUiORicM 

698. The. vnlums of a riycular cylinder is equal to the 
product of its base hij Hh tdlitudc. 



Given the eircnlav I'yli'iLler AJ, having its volume 
denoted by V", its base by B, and its altitude by K. 
To prove r=BXM. 

Proof. Let a prism having a regular polygon for its 
base be inscribed iu the cylinder, and denote the volume of 
the inaeribed prism by T', and its base by B'. 

The prism will have the same altitude, H, as the 
cylinder. 

.-. V'^B'XH. (Why?) 

If the number of lateral faces of the inscribed prism 
be indeflaitely increased, 

V will approach V as a limit. Art. 689, i, 

B' will approach B as a limit. (Whj!) 

And B'X H will approach BXH asd. Hmit (Wh?)) 

But V'^B' X ff always. (Why?) 

.-. V=BX B. (Whys) 

Q. E. a. 

699. Formula for the volume of a circular cylinder. 
By use of Art. 450, 



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CYLINDERS 411 

Proposition V. Theorem 

700. The lateral areas, or the total areas, of hco simi- 
lar cylinderi' of revolution are to each other as the squares 
of their radii, or as the squares of their altitudes; and 
their volumes are to each other as the cubes of their radii, 
or as the cubes of their idfitudcs. 



Given two similar cyliiidura of revolution having their 
lateral areas denoted by S and S', their total areas by T 
and T', their volaraes by T^and P, their radii by R and 
B', and their altitudes by Sand H', respectively. 

To prove S -. H'^T: T^BT- : B'^-^E" -. B'^\ 
and y : y' = E? : R'^ = E'' : S'\ 
_ H+R 
U'~R' R' + R'' 
2 rcitB BXE _R ^,H ^R" ^g" 



Proof. 



Arts, Z'il, 309, 



S' 



r 



I TiR'H' R' X B' 



T,y~rF- 



ir- 



(Wliy?) 



I 7tR' {W + R') B' ff' + E' B'- 



(Whyt) 
(Why f) 

Q. E. B. 



Ex. If a eylindrlcal c 



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CONES 

701. A conical siirface is a sur- 
face giaiierated by ii straight iiin' 
which moves so as constantly to toucli 
a given fixed curve, and constantly 
pass through a given fixed point. 

Tb-tts every shado\r ciist by a ne p 
of iigM iaeonifliil ic foi'iu, tlmt is, is 
by a conical sui'faee of liglit. Hen li 

study ot conical surfaees and solid s a 
portant from the fant that it eone n a 
oases of forces radiating from a neai p n 

702. The generatrix of a 
surface is the moving straight 
AA'; the directrix is the give f d " 
carve, as ABC; the vertex i tixed |ont as 0; an 
element is the generatrix in anv o e ot t po tons as BB'. 

703. Tiie upper and lower nappes oC a conical snrface 
are the portions above and belon' the vertex, respeetivelv 
g,s 0-ABCi\w\ O- A' B'C. 

Usually it is coavenicat lo limit a conical surface to a single nappe, 

704. A cone is a solid bounded by a 
conicai snrface and a plane cuttmg all the 
elements . 

705. The base of a eone is the face 
formed by the cutting plane; the lateral 
surface is the bounding coincal surface; 
the vertex of the cone is the vertex of the i 
conical surface; the elements of the cone 
are the elements of the eonieal surface; cone 

tlie altitude of a cone is the perpendicolar distance from .the 
vertex to the plane of tho base. 




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706. A circular cone is i\ con?, whose base is a circle. 
The axis of a circalm- cone is the line drawn from the 
vertex to the center of the base. 

707. A right circular cone is a c I one 1 e i 
is perpendicular to the plane of the 1 

An oblique circular cone is a cLri, 1 
whose axis is oblique to the base. 

708. A cone of revolution is a co e ge e 
rated by the revolution of a rifjlit t 
about one of its legs as an axis. 

Hence a cone of revolution ai 1 
circular cone are the same solid. ■> 

709. Properties of a cone of revolution inferred im- 
mediately. 

1. The altitude of a con^^ of rrmhiiion U the <uis of the 
cone. 

2. All the elements of a vone ,.f rri'<>]i>/i^n are equal. 

710. The slant height of a cone of revolution is any one 
of its elements. 

711. Similar cones of revolution are uones gciienited by 
similar right triangles revolving about homologous sides. 

712. A plane tangent to a cone i^ a pliine which con- 
tains one element of thu uouu, but wliich does not cut the 
conical surface on being produced. 



Ek, 1. A plane passing througii a taageut lo llic ba.se of a circular 
one and the element drawu through the poiat of coutiict is tnngt-nt to 



Es, 2. ]f a plane is li 
the plane of Iho base i-, t; 



s intersection with 



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713. A pyramid inscribed m 
a cone is a pyramid whose liitci i! 
edges are elements of the coi e 
and whose base is a polygou in 
scribed in the base of the none 

714. A pyramid c i r c u m 

scribed about a cone is a pjia [_ _, 

mid whose lateral faces are tin 

gent to the cone and whose base is a polygon eireumseribed 
about the base of the cone. 

715. Properties of circular cones. By Art, 441 the area 
of a circle is the limit of the area of an inscribed, or of a 
circumscribed polygon, and the circumference is the limit 
of the perimeters of these polygons; hence 

1. The vohmie of a cir<;nlar cone is the limit of the vol- 
ume of an inscribed or eireumseribed pyramid. 

2. The lateral area of a circular cone is the limit of the 
lateral area of an inscribed or circumscribed pyramid. 

716. A frustum of a cone is the \ oi 
tion of tlie cone iuelndcd between th bi e 
of the eoiic and a plane parallel tl 
base. 

The lower base of the frustum 
base of the cone, and the upper 
the frustum is the section made 
plane parallel to the base of the cone. 

What must be the altitude and the lateral surface of a 
frustum of a cone; also the slant height of the frustum ol 
a cone of revolntion ? 




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CONES 415 

Proposition VL Theorem 

717. Every section of ti cone made hy a plane passing 
through its vertex is a triangle. 




Given tlie oone S-AFBQ with a plane passing through 
the Tertes S, and making tlie section SPQ. 

To prove SPQ a triangle. 

Proof. PQ, the intoi-section of tlie base and the cutting 
plane, is a straight line. (Whj!) 

Draw the straight lines SP and SQ. 

Then 8P and HQ must be in the cutting plane; Art. 498, 

And be elements of the (lonical surface. Art. 701. 

,*. the straight lines SP and SQ are the intersections of 
the eonieal surface and the euttiug pliuie. 

.', the section SPQ is a triangle, Art, 81. 

{/or )I is bomuled bij Ihrc: nhviglit lines). 



Ea. What liiiid of triaufjie 
Blade bj a plaxie tiivough the >. 



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BOOK VIII. KOLiD IIEOMETKY 



Proposition VII. Theorj^m 

718. El-cry section of u cirnilar voite made by a plane 
^parallel to the buse is a civl''. 




Given the circular eoue SAB with nph a section made 
by a plane parallel to the base. 

To prove apb a circle. 

Proof. Denote the center of the base by 0, and draw 
the axis, SO, piercing the plane of the section in o. ' 

Through SO and any element, SP, of the conical stir- 
face, pass a plane cutting the plane of the base ia the 
radius OP, and the plane of the section in op. 

In like manner, pass a plane through SO ami SB form- 
ing the intersections OB and oh. 

:. OrWop, and OB]\ob. (Why?) 

.-. A SFO and 8B0 are similar to A Spo and Sbo, 
respectively. Art. 328, 






■' OP 

But OP^OB. (Why J) 

.'. op = ob. {Why t) 

.". oph is a circle. (Wlyt) 

Q. E. ». 

719. Cor. The axis of a circular cone passes through 
the center of every section parallel to the base. 



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CONES 417 

Proposition VIII. Theorem 

720. The lateral arm of a cone of revolution is equal to 
half the product of the slant height by the drcumferenee of 
the base. 




Given a cone of revoltilion having its taterat ares de- 
noted by S, its slant lieight by L, and the cireiiinforeuce 
of ita base by C. 

To prove S^iCXL. 

Proof, Let a regular pyramid bo eireuinscribed about 
the cone. 

Denote the lateral area of the pyramid by S', aad the 
perimeter of its base by P. . 

Then S'^iPXL. Art. 612. 

H the number of lateral faces of the cireumserlbod 
pyramid be indefinitely increased, 

S' will approach S as a limit. Art. 7i5, 2. 

r wilt approach C as a limit. Art. 44i. 

And J P X L will approach J CXL as a limit. Art. 253, 2. 

But S'^i PX L always. (Why?) 

.-. S^^ CXL. (Whyf) 

Q. £. n. 

721. Formulas for lateral area and total area of a cone 
of revolution. Denoting the radius oi' the base by R, 
.S'=i {2 Ttfl X L) :. ti=7TRL. 
Also T=7iiiL +71^- :. T^TcU {L + B). 



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BOOK V!II. SOLID (iEOMETRY 



PkOPOSITIOX IX. TUEOREll 

722. The volume of a circular cone is equal to one-thin 
of the product of its base by its altitude. 




Given a circular cone having its volume denoted by V 
its base by B, and its alfcttude by E. 

To prove V^iBXH. 

Proof- Let a pyramid with a regular polygon (or its 
base be inscribed in the given cone. 

Denote the volume of the inscribed pyramid by F, and 
its base by B'. 

Hence V = ^B'XH. Art. gsi. 

If the number of lateral faces of the inscribed pyramid 
be indefinitely increased, 

V will approach F as a limit, (Why ?) 

B' will approach B as a limit. (Why ?j 

And i B' X H will approach J B X S" as a limit. ( Wby i) 

But V'^i B' X ff always. (Why?) 

r^iBXH. (Why?) 



q. £. i>. 



723. Formula for the volume of a circular c 



Ex. 1. If, in a cone of revolution, if =3 and JJ = 4, find S,r and F, 
Ex. 2. II the altitude of a cone of revolution eijuals the radius of 
the base, what do the formulas for S, T and V become 1 



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CONES 419 

Proposition X. Theorem 

724. The lateral areas, or the total areas, of two simi- 
lar rones of revolution are to each other as the squares of 
their radii, or as the squares of their altitudes, or as the 
squares of their slant heights; and th^ir volumes are to each 
ether as the cubes of these lines. 




Given two similar cones of revolution having their 
lateral areas denoted by S and 8', their total areas by T 
and P, their volumes by Y and V, their radii by jB and 
E', their altitudes by H and H', and their slant heights by 
L and I/, respectively. 

To prove S -. .^'=^T -. T' = IP: R'- = W' -. H'" = J? -. L'"-; 
and V : V' = E? : E''' = H' : H'^ = L^ : L'K 

K V L' + E'' 



Proof. 



(Vfhy?) 



(WhjT) 




725. 

such thf 
triangle. 



(Why?) 
Q. E. D. 

Dek. An equilateral cone is a eone of nivolution 
^. a section through the axis ia au equilateral 



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i2i) BOOS VIII. SOLin GEOJlETRi' 

Proposition XL Theoeem 
726. The lateral area of a frustum of a cone of re.voli 
tioH is equal to one-half the sum of the circumferences of i 
bases multiplied iy Us slant height. 




Given a frustum of a cone of revolution having its 
lateral area denoted by S, its slant height by L, the rarlii 
of its bases by B and r, and the circumfereuces of its bases 
by C and c. 

Toprove S=i (C+c)XL. 

Proof. Let the frustum of a regular pyramid be cireum- 
scribed about the given frustum. Denote the lateral area 
of the eireumscribed frustum by S', the perimeter of the 
lower base by P, and the perimeter of the upper base by p. 

The slant height of the circumscribed frustum is L. 

Hence S'-J {P + p) X L. Art. 643. 

Let the pupil complete the proof. 

(J. E. B. 

727. Formula for the lateral area of a frustum of a cone 
of revolution. S~i {2 TtiJ + 2 tit) L. 

.-. S^n {U + r)L. 

728. Cor. The lateral area of a frustum of a cone of 
revolution is equal to the product of the circumference of Us 
Midsection by its slant height. 



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CONES 421 

Proposition XII. Theorem 

729. The volume of the frustum of a circular cone is 
equivalent to the volume of three cones, whose common alti- 
tude is the altitmle of the frustum, and whose bases are the 
lower hase, the upper base, and a mean proportional between 
the two iases. 



Givea a fnistura of a eircalar cone having its volume 
denoted by V, its altitude by H, the area of its lower base 
by B, and tiiat of its upper base by b. 



To prove V = h S {11+ b + Vb X h). 

Proof. Let the frustum of a pyramid with regular poly- 
gons for its bases be inscribed in the given fmstum. 
Denote the volume of the inscribed frustum by V, and the 
areas of its bases by B' and 6'. 

.-. V' = h S iB' + 6' + l/ff^xT') , (Why t) 

If the number of lateral faces of the inscribed frustum 
be indefinitely increased, V will approach V, B' and V 
approach B and b respectively, and B' X V approach B X 
b, as limits. Art. 719. 

Hence, also, B' + b'+V^B' X b' will approach B -\~ b + 
V'b X J as a limit. Art. 353. 

But V'= iH {B + J'+V-^' X b') always. (Whj?) 

.■. V=iE{B + b + l/fiXb). (Why!) 



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i'12 ];00K VIII. SOIJI) GEOMETRY 

730. Formula for the volume of the frustmu of a circii, 
lar cone. 

F= h if (TtA'- + Tir' + VtiH-'x ni^) . 



Ex. 2. If a eocieal oil-can is 12 iu. iiielii liow much more tin is 
required to make it tliaa to make a airailai' oil-pan in. high ! How 
much more oil wiil it hoid ? 

Ei. 3, The linear dimeneions of i eonieal funnel are three timt-s 
tbose of a similar funnel. How mxieh more tin is required to walie 
the first f How much more liquid will it hold f 

El. 4. Make a similar oompariaon of iiylindrieai oil-tanks. Of 



EXERCISES. CROUP 70 

THBORBMS COKCERNISQ CYLINDERS AND CONES 

El. 1. Any section ot a cylinder of revolution through its axis is 
a rectangle. 

El. 2. On a cylindrical surface only one straight line can be 
drawn through a given point, 

[Sua. For if two straight lines could bo drawn, etc.] 
Es. 3. The intersection of two planes tangent to a cone is a 
straight line through the rertex. 

Ex. 4. If two planes are tangent to a cylinder, their line of inter- 
seetion is parallel to an element ot the eyliuder, 

[SuG. Pass a plane -L to the elements of the cylinder.] 

Ei. 5. If tangent planes be passed through two diametrically 
opposite elements of a circular cone, these planes intersect iu a 
straight line through the Tertes and parallel to the plane of the base, 

Ex. 6. In a cylinder of revolution the diameter of whose base 
equals the altitu<3e, the voiumc equals one-third the product of the 
total surface hy the radius of the base. 



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EXERCISES ON THE CYLINDER AND CONE 423 



Ek. 7. Aeylinder and a cone o! revolutiou have the same base 
Bnd the same altitude. Ficd tbe ratio of their lateral surfaoes, and also 
of their voluroes. 

Ex. 8. It an equilateral triangle whose Bide U a be revolved iiljout 
ooe of ita Bides as an asis, find (he area generated in terms of a. 

Ex. 9. If a rectangle whose aidea are a and b be revolved first 
about the aide a as an axis, and then about the side b, find the ratio of 
the lateral areas generated, and also of the volumes. 



Ex. 10. The baaoe of a cylinder 
ooacentric. The two solids have th 
ot the base of the cone is twice the < 
def. What kind of Hue is the iute 
and how far is it from the base 1 



it a cone of revolution are 
attitude, and the diameter 
er of the base of the cjlin- 
a of their lateral surfaoea, 



Ejc. 11. Determiue the same when the radius of tha «■ 
imea the radius of the cylinder. Also when r times, 

Ex. 12. Obtain a formula in terms o£ r for the volume of the 
'rustum of an equilateral cone, in which the radius of the upper base 
a r and that of the lower base is 3r. 



Ex. 13. A regular hoxagon whose aide 
oual through the center as axis. Find, i 
and volume generated. 



folvea about a diag- 



Ex. 14. Find the locus of a point at a giv 


en distance f 


straight line. 






Ex. 15. Find the locus of a point whos 


e distance f 


line is in a given ratio 


to ita distance froc 


a a fixed pis 


dioular to the line. 






Ex. 16. Find the lo 


us of all straight 1 


nea which m 


angle with a given line 


at a given point. 




Ex. 17. Find the lo 


cus of all straight 1 


nea which m 


angle with a given plan 


e at a given point. 





Ex. 1 8. Find the locus of all points at c 
tanee from the surface of a given cylinder of 



n the surface of a gvv 




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JK vni. SOLID GEOJIIM'KY 

EXERCISES. QKOUP 71 

)N'Cf.KN'IXG Till'; l'VLlSr>F.[l AND CONE 

Ex. !. Through a given element o( a tireular eyliader, paaa v 
plane l.augent to tbe ejlicder. 

Ex. 2. Through a given element of a circular ciine, pass a plane tniij^Li 



Ex. 3. About a given circular oylind 
s regular polygon for it^ base. 

Ei. i. Through a giren point outside a circuUr cylimier, paas a 
plane tangent to the cyiinder. 

Ex. B. Through a given point outside a given elnmlar eone, pas3 
a plane tangent to tbe cone. 

[St'ti. Through tbe vertei of tlio cone and the given point pass a 
line, and produce it to meet the plane of the base.] 

Ex. 6. Into what eegiuents must the altitude of a coneof revoliitirjn 
he divided by a plane parallel to the base, in order that the volume of 
the cone be bisected? 

Ex. 7. Divide the lateral surface of a given cone of revolution into 
two equivalent parts by a plane parallel to the base. 

Ex. 8. If the lateral surface of a cylinder ot revolution be ont 
along one element and unrolled, what Movt of a plane figure is formed f 

lience, out of cardboard eonatruot a cylinder of revolution with 
given altitude and given eireumterenee. 



Ex. 9. If the 


literal ^ 


uitice 


of ac 


one of 


revolution be 


out along 


one element and 


umoHed 


what 


sort ol 


: a. plai 


16 figure is formed ! 


Hence, out ol 


taidboa 


li toi 


JStiuet 


a com 


e of revohition 


ot given 


slant height 














Ex. 10. C(n= 


trmtan 


e^nlla 


teril CI 


TOO ont 


of pasteboard. 




Ex. 11, eout 


itruit a 1 


luatliu 


. Of a< 


■one of 


revolution out 


, of paste- 


board. 















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Book IS 
THE SPHERE 

731. A sphere is a solid bounded by a surface all 
points of which are equally distant from a point withiu 
called the center. 




732. A sphere may also be defined as a solid generated 
by the revolution of a semicircle about its diameter as an 
axis. 

Some oE the properties of a epheio may be obtained more readily 
from one o( the two deflcitions given, and some from the otliec. 

A sphere is named by naming tiie point at its eecter, or by naming 
three or more points on its surfaae. 

733. A raduis of a sphere is a line drawn from the 
center to any point on the surface, 

A diameter of a sphere is a line drawn through the 
center and terminated at each end by the surface of the 
sphere. 

734. A line tangent to a sphere is a line having bnt one 
point in common with the surface of the sphere, however 
Jar the line be produced. 

(425) 



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4::f) BOOK IX. SOLID GEOJffiTRY 

735. A plane tangent to a sphere is ;i plane having bu 
one point in common with tlifi surface of the sphere, how- 
ever far the plane be produced. 

736. Two spheres tangent to each other are spheiv-- 
whose surfaces have one poiut, and only one, in common. 

737. Properties of a sphere inferred immediately. 

1. All radii of a sphere, or of equal spheres, are equal. 

2. All diameters of a sphere, or of equal spheres, are. 
equal. 

3. Two spheres are equal if their radii or their diame- 
ters are equal. 



PROi'OSiTioN I. Theorem 
738. A section of a sphere, tnade by a plane is a circle. 




Given HI 1 1 1.) (i I id PCD a section made by a plane 
cutting the sphere. 

To prove that FGD is a circle. 

Proof. From the center 0, draw OA X the plane of 
the section. 



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THE SPHERE 427 

Let Che B. fixed point on the perimeter of the section, 
and P any other point on this perimeter. 
Drawee, AP, OC, OP. 

Then the A OAP and OAG hvb rt. A. Art. 505. 

OP=OG. (Whj!) 

OA^OA. (Why!) 

A OAP=A OAC. (Whjf) 

.-. AP=AC. (Wiiy t) 

But P is any poiut on the perimeter of the section PCD. 

:. every point on this perimeter is at the distance AO 

from A. 

.". PCD is a circle with center A. Art. 197. 

q. E. s. 

739. Cor. 1. Circles which are se.ctions of a sphfre 
wade by planes equidistant from the center are equal; and 
conversely. 

740. CoE. 2. Of two circles on a sphere, the one nuide 
by a plane more remote from the center is smaller; and 
conversely. 

741. Dep. a great circle of a sphere is a circle whose 
plane passes through the center of the sphere. 

742. Dep. A small circle of a sphere is a circle whose 
piane does not pass through the center of the sphere. 

743. Dep. The axis of a circle of a sphere is the 
diameter of a sphere which is perpendicular to the plane of 
the circle. Thus, on figure p, 42fi, BB' is the axis of PCD. 

744. Dep. The poles of a circle of a sphere are the 
"istremities of the axis of the circle. Thus, P and B', of 
■igure p, 426, are poles of the circle PCD. 



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428 BOOK i\. SOLID f;EOMETKY 

745. Properties of circles of a sphere inferred imiiif 
diately. 

1. 2'Ae axiK of a circle of a sphere pnxses through tl 
center of the circle; and conversely. 

2. Parallel circles have the same axis and the same poln 

3. All great circles of a sphere are equal. 

4. Every great circle on a sphere bisects the sphere ai, 
its surface. 

5. Two great circles on a sphere bisect each other. 

For the line of intersection of the two planes of th 
circles passes througli the center, and hence is a diametf' 
of each circle. 

6. Through two points (not tite extremities of a diameio 
on the surface of a sphere, one, and only one, great circle ch 
he passed. 

For the plane of the great circle nmat also pass throng 
the center of the sphere (Art. 741), and through thro 
points not in a straight line only one plane can be passe 
(Art. 500) . 

7. Through any three points on the surface of a sphen 
not )rt the same plane icith the center, one small circle, a/t 
only one, can he passed. 

746. Def. The distance between two points on the sui 
face of a sphere is the length of the minor arc of a grt-a 
circle joining the points. 



Ex. 1. If the raiii.is of a sphere 


is 13 in-, find the vadlua c 


circle on the sphere made by & plane 


at a distance of 1 ft. from 


center. 




Ex.2. Wliat geographical eirclea 


00 the earth's surface are gn 


and what small circles t 




Ex. 3. What is the largest nuiabt 


ir of points in which two cir 


ou tlie surface of a aphevo can iutersi 


Eictf Why? 



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THE SPIIEKE 41!y 

Proi'OSition II. Theorem 
747. All points in the circumference of a circle of a 
sphere are eqvnlhi lU^tnut from each pole of the circle. 




Given ABC a circle of a sphere, and P atid P' its poles. 
To prove the arcs FA, PB, PC equal, and ares F'A, 
r'B, P'G equal. 

Proof. Draw the chords PA, PB, PC. 
The chords PA, PB and PC are equal. Art. 518. 

,■. area FA, PB and PC are equal. Art. 313. 

la like manner, the arcs F'A, F B and FC may be 

Q. E. B. 



proved equal. 



748. Def. The polar distance of a small circle on a 
sphere is the distance of any point on the circumference of 
the circle from the nearer pole. 

The polar distance of a great circle on a sphere is the dis- 
tance of any point on the eircumfereuce of the great circle 
from either pole. 

749. Coil. The polar distance of a great circle is the 
quadrant of a great circle. 



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430 BOOK IS. SOLID GE0\£TE¥ 

Proposition III. Theoreji 

750. // a point on the surface of a sphere is at a quaii 
rfinf's dhhmce from two other poinl.i on the. surface, it i. 
the pole of the great circle through those points. 




Given PB and PC quadrants on the surface of the 
spliere 0, and ABC a, great circle through B and C. 

To prove that P is the pole of ABC. 

Proof. From the center draw the radii OB, OC, OP. 

The arcs PB and PC are quadrants. (Why t) 

.■- ^ POB and POO are rt, A. (WhyT) 

.-. PO ± plane ABC. (Why?) 

.". P is the pole of the great circle ABC. (Why i) 

Q. 1. D, 

751. Cor, Through two given points on the surface of 
a sphere to describe a great circle. 

Let A and B be the given points. 
From A and B as centers, with a 
quadrant as radius, describe arcs 
on the surface of the sphere inter- 
secting at P. With P as a center 
»nd a quadrant as a radius, describe a great circle. 




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THE SPHEliE 

Proposition IV. Theorem 
752. A plane perpendimlar to a radius at its extre 
is tangent to the sphere. 




Given the sphere 0, and the place MN J. the radius OA 
of the sphere at its extremity A. 

To prove MN tangent to the sphere. 

Proof. Take P any point in plane MN except A. Draw 
OP. Then OP>OA. (Whj-f) 

.■. the point P is outside the surface of the sphere. 

But P is any point in the plane MN except A. 
.: plane MN is tangent to the sphere at the point A, 
(for every point in Vie plane, except A, U milstde the surface fif t!ie sphere). 
Art. 73&. a, B. D. 

753. Cor. 1. A pkinc, or a line, which is tangent to a 
Sphere, is perpendicular to the radius drawn to the point of 
<:ontact. Also, if a plane is tangent to a sphere, a perpendicii- 
far to the plane at its point of contact passes through the center 
of the sphere. 

754. Cor. 2. A straight line perpendicular to a radius 
''J a sphere at its extremity is tangent to the i^phere. 

755. Cob. 3, A straight line tangent to a circle of a 
Sphere, lies in the plane tangent to the sphere at the point of 



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4:62 BOOK IX. SOLID GKo:\iT:T]iy 

756. Cob. 4. A straight line drawn in a tangent plan 
and through the point of contact is tangent to the sphere : 
that point, 

757. Cor. 5. Two straight lines tangent to a sphere , 
a given point determine the tangent plane at that point. 

758. Def. a sphere circumscribed about a poiyhedro 
is a sphere in whose surface lie all the vertices of tli 
iiolyhedron. 

759. Df.p. a sphere inscribed in a polyhedron ij; 
sphere to which all the faces of tlic polyhedron are tangent 



PROPOSITION V. Problem 

760. To circumscribe a sphere about a give 
hedron. 




Given the tetrahedron ABCD. 

To circumscribe a sphere about ABCD. 

Construction and Proof. Construet E and F the centers 
of circles circumscribed about the A ABC and BGD, re- 
spectively. Art. SSfi. 

Draw ES X plane ABC and FK X plane BCD. Act. 5U. 

Draw FG and FO to G the midpoint of BG. 



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THE SPHERE Add 

Then EG and FG are X BC. Art. ii3. 

.-. plane EOF 1 BC. Art. 509. 

.-. pljine EOF ± plane ABC. Art, 555. 

.-. SH lies in the plane FOE. Art. 558. 

In like manner FK lies in the plane F6E. 

The lines EG and FG are not ll, 

[for Ihfy nieel in the point G), 
:. the lines Bff and -F£: are not !1. Art. 122. 

Hence ES innst meet FK in some point 0. 
But EJ/^ is the locus of all points equidistant from A, 
B and C\ and F^ is the loeus of all points equidistant 
from B, C and Z>. Art, 520. 

,'. 0, which is in both EH and FK, is equidistant from 
A, B, Oand D. (Whyl) 

Hence a spherical surface constructed with as a center 
and OA as a radius will pass through A, B, G and D, anil 
form the sphere required, q, e. f. 

761. Cor. 1. Four points not m Ike same plane deter- 
mine a sphere. 

762. Cor. 2. The four perpendiculars erected at the 
'"'enters of the faces of a tetrahedron meet hi a point. 

763. Cor. 3. The sije planes perpendicular to the edges 
of a tetrahedron at their midpoints intersect in a point. 

764. Dep, An angle formed by two curves is the angle 
formed by a tangent to eaoh curve at the point of inter- 
section. 

765. Def. a spherical angle is an angle formed \iy 
two interseetiutr avns of great circles on a sphere, and 
l>euee by tangents to these arcs at the point of intersection. 



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34 BOOK IX. POLID r.EOMKTRY 

PkOPOSITION VI. PltOHLEM 

766. To inscribe a sphere in a given ieffihrilrtm. 




Given the tetraheilion ABOD 

To inscribe a spheie m ABC1> 

Coaatruction and Proof. BibPct the dihoilral angle D- 
AB-C hy the plane OAB; similarly bisect tlie diliedral A 
■whose edges are BG and AG hy the planes OBC and OAG, 
respectively. 

Denote the point in whieh the three bisecting planes 
intersect by 0. 

Every point in the plane OAB is equidistant from the 
faces DAB and GAB. Art. 562. 

Similarly, every point in OBC is ec|uidistant from the 
two faces intersecting in BG, and every point in OAG is 
equidistant from the two faces intersecting in AG. 

.: is equidistant from all four faces of the tetra- 
hedron. Ax. 1. 

Hence, from as a center, with the J. from to any 
one face as a radius, describe a sphere. 

This sphere will te tangent to the four faces of the 
tetrahedron and .'. inscribed in the tetrahedron. Art. 759. 



767. OoR. The planes bisecting the s 
of a tetrahedron meet in one point. 



: dihedral angles 



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THE SPHERE 

Proposition VII. Problem 
768. To find the radius of a given material sphere. 




Given the material sphere 0. 

To construct the radius of the sphere. 

Constnictioa. With any point P (Fig, 1) of the surface 
of the sphere as a pole, describe any" convenient circum- 
ference on the surface. 

On this eireumferenee take any three points A , B and C. 

Construct the A ABC (Fig. 2) having as sides the three 
chords AB, BC, AC, obtained from Fig. 1, by use of the 
compasses. Art. 283. 

Circumscribe a circle about the A ABC. Art. 286. 

Let KB be the radius of this circle. 

Construct (Fig. 3) the right A kpb, having for hypot- 
enuse the chord pb {Fig. 1) and the base ii. Art. 384. 

Draw bp' ± hp and meeting pk produced at p'. 

Bisect /jp' at 0. 

Then op is the radius of the given sphere. 

Proof- Let the pupil supply the proof. 

Q. E. F. 



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436 BOOK IX. SOLID GEOTilKTRY 

Proposition VI]!. Theorem 

769. The intersection of itvo spherical surfaces is the 
circumference of a circle whose plane is perpmdicularHo the 
line joining the centers of the spheres, and whose center is i)i 
that line. 




Given two intersecting © and 0' wtiinh, by rotation 
about tlie line 00' as an axis, generate two intersecting 
splierical siirfaces. 

To prove that tlie intersection of the spherical surfaces 
is a ©, whose plane ± 00', and whose center lies in 00'. 

Proof, Let the two circles intersect in the points P and 
Q, and draw fhe common chord PQ. 

Then, as the two given ® rotate about 00' as an axis, 
the point P will generate the line of intersection of the two 
spherical surfaces that are formed. 

But PR is constantly 1 OO". Art. 241, 

.', PB generates a plane J. 00' Art, 510, 

Also PR remains constant in length. 

,', P describes a circumference in that plane. Art. 197. 

Hence the intersection of two spherical surfaces is a O , 
whose plane X the line of centers, and whoso center is in 
the line of centers. 

Q. E. B. 

Tlie abovft demonstration is an illustration of the use of the second 
definition of a sphere (Art. 732). 



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THE SPHEKE 437 

Peoposition IX. Theorem 

770. A spherical angle is measured by the arc of a great 
circle described from the vertex of the angle as a pole, and 
included between its sides, produced, if necessary . 




Given /.BAG a spherical angle formed by the intersec- 
tion of the arcs of the great circles BA and €A, and BO 
an are of a great circle whose pole is A. 

To prove /. BAG measured by are BG. 

Proof, Draw AI> tangent to AB, and AF tangent to 
AC. Also draw the radii OB and OC. 

Then Al> X AG. Art. 230. 

Also OB L AO iJorABisaqm-iranl). 

:. OB WAD, (Why?) 

Similarly OGWAF. 

:. ^B0(J= IDAW 
But Z BOG is measured by arc BG. 
:. IDAF, that is, LBAG, is measured by t 



Art. 538. 

Art. 257. 

■■C BG- 

(Why?) 
Q. E. 9. 



771. Cor. A spherical angle is equal in the plain 
ofthv dihtdral angle formed hij (he planer of Us sides 



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4r,S BOOK IX. HOLIU liEOJIETUY 

SPHERICAL TRIANGLES AND POLYGONS 

772. A spherical polygon is a portion of the 
a sphere boumlyd by three or more 
arcs of gi'eat circles, as ABGD. 

The sides of the spherical polygon 
are the bounding arcs; the vertices 
are the points in which the sides in- 
tersect; the angles are the spherical 
angles formed by the sides. 



773, A spherical triangle is a. splieriea! poiygon of three 
sides. 

Spherical triangles are classified in the same way as 
plane triangles; viz., as isosceles, equilateral, scalene, 
right, obtuse and acute, 

774. Relation of spherical polygons to polyhedral angles. 
If radii be drawn from the center of a sphere to the ver- 
tices of a spherical polygon on its surface (as OA, OB, 
etc., in the above figure), a polyhedral angle is formed at 
0, which has an important relation to the spherical poly- 
gon ABGD 

Bach face angle of the polyhedral angle equals {in yium- 
her of degrees contained) the corresponding side of the spheri- 
cal polygon; 

Dach dihedral angle of the polyhedral angle equals the 
corresponding angle of ike spherical polygon. 

Hence, corresponding to each property of a polyhedral 
angle, there exists a property of a spherical polygon, and 
conversely . 



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THE SPHERE 4J9 

Ilenoe, also, a trihedral angle and its parts correspond 
to a spkfricnl triangle and its parts. 

Of the common proportieB of a polyheilral angle and a epherical 
polygon, some are discovered more readily from the one figure and 
some from the other. In general, the spherical polygon ia simpler 
to deal with than a polyhedral angle. For instance, if a triheUral 
angle were drawn with the plane angles of its dihedral angles, nine 
lines would be used, forming a complicated tigure in solid apace; 
whereas, the same magnitudes are represented in a apherieal triangle 
by three lines in an approximately plane figure. 

On the other hand, the spherieal polygon, beeauao of its lack of 
detailed parts, is often not so suggestive of properties as the poly- 
hedral angle. 



Proposition X. Theorem 

775. The sum of tu'o sides of a spherical triangle is 
greater than the third side. 




Givea the spherical triangle ABC, of whieh uo side is 
larger tliaii AB. 

To prove AC-^ BO AB. 

Proof. From the center of tlie sphere, 0, draw the radii 
OA, OB, OC. 

Then, in the trihedral angle O-AHG, 

/LAOC. + IBOV > AOB. Art. 582. 

.-. AV-^ BO > AH. Art. 774. 

Q. E. 5. 



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440 BOOX IX. SOLID CKOUETllY 

776. CoK. 1. Any side of a spherical iriautjleis gfealn- 
thin the. difference betiveen the other two mles. 

777. Cor. 2. TJie shortest path ietween two points on 
the surface of a sphere is the arc of the great circle joining 
those points. 

For any other path between" the two points may be 
made the limit of a series of arcs of great oircies connect- 
ing successive points on the path, and the sum of this 
series of arcs of great circles connecting the two points is 
greater than the single arc of a great circle connecting 
them. 

Proposition XI, Theorem 

778. The smn of the siiits of a spherical polygon is less 
than 360°. 




Given the spherical polygon A BCD. 
To prove the sum of the sides of ABCD < 360°. 
Proof. From 0, the center of the sphere, draw the radii 
OA, OB, OC, 01). 

Then lAOB-\- IBOG+ lOOI>-\- IDOA < SCO". 

(Why T) 
.-. AB -^ B(] + CD -^ DA <360°. Art, 77*. 

Q. E. I>. 



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SPHERICAL TKIANGLES 



441 



779. Def. The polar 
triangle of a given triangle 
is the triangle formed by 
taking the vertices of the 
given triangle as poles, and 
deserihing arcs of great cir- 
cles. (Hence, if each pole 
be regarded as a center, the radius used iu describing each 
arc is a quadrant.) Thus A'B'C'is the polar ti'iangle of 
ABC; also B'E'F' is the poiar triangle of BEF. 




Proposition XII. Theorem 

780. If one spherical triangle is the polar of another, 
then the second triangle is the polar of the first. 




Given A'B'C the polar triangle of ABC. 
To prove ABC the polar triangle of A'B'C . 
Proof. B is the pole of the arc A'C . 

:. arc A'B is a quadrant. 
Also C is the pole of the arc A'B'. 

.'. arc A'C is a quadrant. 
.*. A' is at a quadrant's distance from both B and C. 

.*. A' is the pole of the arc BC. Art. 750. 

Iq like manner it may be shown that B' is the pole of 
AC, and C the pole of AB. g. E, 9. 



Art. 779. 
(Why t) 

(Why?) 
(Why?) 



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442 I'.OOK !X. SOLID aEOlIETUY 

Phoposition XIII. Theorem 

781. Ill ii sjihfrical triangle <iiid its polar, each angle of 
one Irianglc is the siipphmeiii of the side opposite in the 
other li-iangle. 




Art. tso. 
(Whyf) 
(Why!) 



Given the polar ^ ABC ami A'B'C with the sides of 
^£(7 denoted by a, b, c, and the sides of A'B'C denoted 
by «', h', c', respectively. 

To prove A + «' = 180°, B + 6'=180°, C' + c'=180°, 
A' + a= 180°, B' + b= 180°, C + c - 180°. 

Proof. Produce the sides AB and AG till they meet 
B'C in the points D and F, respectively. 

Then B' is the pole of AF :. arc B'F=90' 

Also C is the pole of AD :. arc CD -90' 

Adding, B'F + C'I>=180°. 

Or B'F + FC + .BF^ 180°. Ax. 6- 

Or 7J'C' + 7H'-180°. 

But B'C'==a', and /ii'' is the measnre of the A A. Art, 770. 
.-. A + a' -180°. 

In like manner the other supplemental relations may be 

proved as specified. 

Q. a. B. 

782. Def. Supplemental triangles are two spherical 
triangles each of which is the polar triangle of the other. 

This new name for two polar triangles is due to the 
property proved in Art. 781. 



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SPHERICAL TRIANGLES 443 

Proposition XIV, Theorem 

783. The sum of the angles of a spherical triangle is 
greater than 180°, and less than 540°. 




Given tliG spherical triangle ABC. 
To prove J + S + C > 180" and < S40°. 
Proof. Draw A'B'C, tlie polar triangle of ABO, and 
denote its sides by a', S', c'. 

Tiien A + a' - 180" 1 

B -\- I' = 180" > Art. m. 

C + c' = 180" ' 
.-. A -r B + C + a' + b' + c' = 540°. . . (1) Ax, 2. 
Hut ^a' + b' + c' < 3ilO' Ai-i. -7R. 

j „' 4- i' + c' > 0= 
Subtracting eacb of these In tnni from (1), 

A + B + C > 180° and < 540". ak. ii. 

(I. E. B. 

784. COH. A spherical triangle may have one, two or 
three right angles; or it may have one, two or three obtuse 
angles. 

785. Def. a birectangular spherical triangle is a splier- 
wal triangle containing two right angles, 

786. Dei\ a trirectangular spherical triangle is a 
spherical triangle containing three right angles. 



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444 



HOOK IX. SOLID GEOMETIiV 



787. GoR. The surface of it sphere may he dimded into 
fight inrectaiigiilar spherical triangles. For let three planes 
X to each other be passed through the center of a sphere, etc. 



788. Def. The spherical excess of a sphei-lciil t.ria 
is the excess. of the sum of its angles over 180°. 



igle 



789. 

whieh Y 



Dep. Symmetrical spherical triangles are triangles 
ive their parts eqnai, but arranged in reverse order. 




Three planes passing through the isenter of a sphere 
form a pair of symmetrical spherical triangles on opposite 
sides of the sphere (see Art. 580), as & ABC and A'Ji'C 
of Fig. 1. 

790. Equivalence of symmetrical spherical triangles. 
Two plane triangles which have their parts equal, but ar- 
ranged inreverse p' 
order, may be a^ /\ 
made to coincide / \v /^ \ 
by lifting up one / x. ^ ^ 'X X ^-' 
triangle, turning 
it over in space, and plaeiug it upon the other triangle. 

But two symmetrical spherical triangles cannot be made 
to coincide in this way, because of the curvature of a 
spherical surface. Hetiee the equivalence of two sym- 
metrical spherical triangles must be demonstrated in some 
indirect way. 



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SPHERICAL TEIANGLES 443 

791. Property of symmetrical spherical triangles. Two 
isosceles symmetrical spherical triangles are equal, for they 
can be made to coincide. 

Proposition XV. Theorem 

792. Two symmetrical spherical triangles are equivalent. 




Given the i 
formed by plane 
(See Art. 789.) 

To prove 

Proof. Let 
the points A, B, 

Also draw 
great ®. 

Abo P' 



lymmetrical spherical A ABC and .I'R'C, 
s passing through 0, the center of a sphere. 

A AnCOA A'B-C. 
P be the polo of a small circle passing through 

Draw the diameter POP'. 
PA, PB, PC, P'A', P'B', P'C, all arcs of 



PA = PB t= PC. 
--PA, P'B' ^ PB, P'C = PC. 



Ax. 1 



Similarly 

And 

Adding 



In case the 
A'B'C, V-X the 



, ,'. P'A' = P'B' =P'C'. 
J and P'A'B' are symmetrical isosceles 

A PAG^^FA'C. 

A PBC=^ PB'C, 

l^PAB-+APAC+APBC 

^AP'A'B'+/^FA'C'-\-AP'B'C'. kx. 

AABC^AA'B'C. Ax. 

polos /'and 7" fall outside the ^ ABC iiii 

nuni! simply the demoiistratiou. q. e. b. 



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BOOK IX. SOLID r.EOMK'niY 



i'Hiii'o^rnoN XVI. Tii 



793. On the same sphe? 
{ingles are equal, 

I. If two sides and the included angle of one are equal to 
liro .sides atid the included angle of the oilier; or 

JI. Ifiivo angles and the inehtded side of one a 
io two angles and the included side of the other, 

the corresponding I'quil parts being arranged in the saw/; 
order in each case. 



r on equal spheres, ivn ;»■;. 



ire equitl 




I. Given the spheriRai A ABC and J>EF, in which 
AC=i>F, CR = FE, and ZC=IF. 

To prove A ABC= A DEF. 

Proof. Let the pupil supply the proof (see Book J, 
Prop. YD- 

II. Given the spherical & ABC and DEF, in whieh 
Z(";=ZF, lJi = /.E, unA CB^FE. 

To prove A ABG^ A J)EF. 

Proof. Let the pupil supply the proof (see Book I, 



Ex. 1. 

tre 12 in. 


If the 
and 3 


Ex. 2. 

15 inciiea 


The t£ 
In dmm 



lineof centers of two spl: 
ill,, liow ai'e tho spheres 

auk on a motor car is a crylltider l!6 inches long and 

leter. How many gallons of gasolene will it hold 1 

equilateral cone, find the ratio of the lateral area to 



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SPHEEICAL TRIASGLES 



Peoposition XVII. Theorem 

794. On the same sphere, or on equal spheres, two tri- 
angles are symmetrical and equivalent, 

I. If two sides and the inchtded angle of one are equal to 
two sides and tlie included angle of the other; or 

II. If two angles and the inrAuded side of one are equal 
to two angles and the included side of the other, 

the cotfesponding equal parts being arranged in reverse 
order. 





I. Given the spherical & ABC and DSF, in which AB 
= DE, AC = DF, iind £A = £D, the eorrespomling parts 
being arranged in reverse order. 

To prove A ABC symmetrical with A DEF. 

Proof. Construct the /MVE'F' symmetrical with ADEF. 

Then A ABC may be made to coincide with A D'E'F', 
Art, 703. 
{haring two sides and the hiclmh-d L etiuat ami arraiigeil in llie 
>ameor,hy). 

But A D'E'F' is symmetrical with the A DEE. 
:. A ABC, which coincides with A D'E'F, is symmet- 
rical with A hEF. 



ir. The second yiirt of the theorem 
same way. 



; proved in t! 
I), E. D. 



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BOOK IX. SOLID GEOMKTKV 



Proposition XYIII. Theorem 

795. If two triangles oti Ike same sphere, or equal 
spheres, are mnhtally equilateral, they are also imitnally 
equiangular, and therefore equal or symmetrical. 



Given two mutually equilateral spherical A ABC and 
A'B'C oil the same or on equal spheres. 

To prove A ABC and A'B'C equal or symmetrical. 

Proof. Prom and 0', the centers of the spheres to 
■which the given triangles belong, draw the radii OA, 
OC, O'A', O'B', O'C. 

Then the face A at — corresponding face A at (y. 

Art. 774. 

Hence dihedral A at = corresponding dihedral A at 0', 

Art. 584 

,'. A of spherical A .4 £(7= homologous A of spherical 
/\ A'B'C. Art. 774, 

.". the A ABC and A'B'C are equal or symmetrical 
ficcording as their homologous parts are arranged in the 
same or in reverse order. Art. 789. 

796. Note. Theconditions in Props. XVlaad SVIII which make 
two sphei'iea! triangles equal aro the same as those which malse 
two plane triangles equal Hence many other propOBitions occur in 
Bpherieal geometry which are identical with correspondlag proposi- 
tions in plane geometry. Thus, many of the construction problems of 
epherioal geometry are Bolvd in. the sama way as the corresponding 
construction problems m plane jfeomelryi as, to bisect a given 
aogle, etc. 



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'SPHEBICAL TEIANGLES 449 

Proposition XIX. Theorem 

797. If two Iriangles on the same sphere are mutually 
equiangular, they ore idno mutually equilateral, and there- 
fore equal or symmetrical. 



Given the muhially equiangular spherical ii Q and Q' 
on the same sphere or on equal spheres. 



iitiiaOy equilateral, and 



To prove that Q and Q" are i 
therefore equal or symmetrioal. 

Proof. Oc.nHtruct P and P' the polar & of Q and Q', 
respective b . 

Then A P aud P' are mutually equilateral. Art. 7S1. 

.". iSv P and P' are mutually equiangular. Art. 795. 

But Q is the polar A of P, and Q' of £". Art. 780. 

.". A Q and Q' are mutually equilateral. Art. 78i. 

Henee Q and Q' are equal or symmetrical, according as 
their homologous' parts are arranged iu the same or in 
:.'everse order. Art. 789. 

Q. E. B. 

798. CO!E. If tu-o mutualhj equiangular tiianijtes are 
0)1 unequal uplfren, their correspomUng sides have the same 
ratio as the radii of their respective spheres. 



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450 liOOK [X. SOLID CEOMETliV 



I'HOFOSITION XX. TlIEOlJEM 

799. Jii an isosceles spherical triangle the angles opp,, 
site the eqnul sides are equal. 



Given the spherit-a! A ABC in whii^li An=-AC. 
To prove Z7J=ZC. 

Proof, Draw an arc from the vertex .1 to i>, the mid- 
point of the base. 

Let the pupil suppb' the roraaiiider of the proof. 

I'KorosiTiON XXI. Theorem (Conv. of Prop. XX) 
800. Jftivo angles of a spherical triani/le are equal, the 
sides opposite these angles are equal, and the triangle is 
isosceles. 



Given the spherical A ABG in which ZB= Z 0. 
To prove AB^AC. 

Proof. Construct A A'B'G' the polar A of ABC. 
Then A'C' = A'B'. Art 

.-. ^6"= Z7J'. Art 

.-. AB = AG. Art 



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SI'IIEEICAL TRIANGLES 451 

Proposition XXII. Theorem 

801. 2m any spkt^rical triangle, if two angles are tm- 
equaJ, the sides opposite these angles are unequal, and the 
greater side is opposite the greater angle, and Conversely. 




Given the spherical A ABC in which ,^. JiAO is greater 



To prove liC > BA.  

Proof. Draw the iirc AD making IPAO pqw.\\ lo 10. 
Theu JiA = DC. Art,. 800. 

To each of these equals add the are HI). 

:. BD+ J)A^TiT>-\- T>(\ or liC. (Why?) 

But ill A BBA, BI) + T)A > BA. (Wl.yT) 

.-. BO > HA. Ax. 8, 

Let ihe popii ijrnve the eonver.se by the indirect method 

see Art. 106). 

Q, £, D. 



Ex. 1. Bisect a givi^n spherical aHgle. 

Es. 2. Bisect B given are of a great circle oii a sphei'f- 

(■'lii^il to a given spherical angla on tlie same sphere. 

Ex. 4. Fimi (lie ioeiia ot the ccoterB o( tbe oirolea o£ a epher 
foi'iiifd by pianos perpendloulac to a given ilinmeter of tlip give 
'pheru. 



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i'y2 



ROLTD GEOMElliY 



SPHERICAL AREAS 

802. Units of spherical surface. j\ 
may be measured in tonus of, eirlicr 

1. The customary units of arm, as 
square foot, etc., or 

2. Spherical degrees, or spherids. 

803. A spherical degree, or spherid, is one-ninetieth 
part of one of the eight trireetangnlar triaugles into which 
the surface of a sphei-o may be divided (Art. 787), or -^is 
part of the surface of the entire sphe 

A solid degree i 



spherical surface 
a square inch, a 



ie-nicet!eth part of a tri rectangular angle (se 



lU). 



I sphere 



804. A lune is a portion of the surface of i 
bounded by two semicii-eumferences 
of great circles, as PBPC of Fig. 1. 

The angle of a lune is the angle 
formed by the semieircumfevences 
which bound it, as the angle BPG. 

805. A zone is the portion of the 
surface of the sphere bounded by 
two parallel planes. 

A zone may also be defined as the sur- 
face generated by an arc of a revolving 
semicircumfetenpe. Thus, if QFQ' {FIb- 2) 
generates a sphere by rotating about QQ", 
its diameter, any arc of QFQ', as ISF, 
generates a zone. 

806. A zone of one base is a zone 
one of whose boiitiding planes is 
tangent to the sphere, as the zone 
generated by the arc QtJ of Pig. 2. pig. -.-. 

807. The altitude of a zone is the perpendiuulat- tiis 
tance between the bounding planes of the zone. 

The bases of a zone are the circumferences of the circles 
of the sphere formed by the bounding planes of the zone. 




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Spherical areas 4yd 

Proposition XXIII. Theorem 

808. The area generated hy a straiglit line revolving 
about CHi axis in its plane is equal to the projection of the 
line upon the axis, mtiUipUed by tM circumference of a circle 
whose radius is the perpendicular erected at the inidpoint of 
the line and terminated by the axis. 




Given AB and X¥ in tlie same plane, CD the projection 
of AB on XY. PQ the ± bisector of AB; and a surface 
generated by the revolution of AB about XY, denoted as 
"area ATi." 

To prove area AB^ CD X 2 nPQ. 

Proof. 1. In genera!, the surface generated by AB is 
tlie surface of a frustum of a cone (Fig. 1). 

,-, area AB=AB X 2 tiFR. am. 72S. 

Dvaw AFLBD, then 

A ABF and PQE are similar. Art. 32s. 

.-. AB: AF^PQ: PR. (Why!) 

.-. AB X P7^ = AZ'■'X PQ, or 071 X PQ. (Why!) 

Substituting, area AB= CBX1 nPQ. Ak. 8. 

2. If ABWXY {Pig. 2), the surface generated by AB 

is the lateral surface of a cylinder. 

.-. area AB- Ci> X 2 tiPQ. Art, ddl. 

■S. If the point A lies in the axis XY (Fig. 3), k't the 
pupil show that the same result iw obtained. q. e, d, 



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4j4 book IX. SOLID GEOJIETRY 

PiiO POSITION" XXR'. Theorem 

809. The area of the surface of a sphere is equal io fht 
prodiin of the diameter of the sphere by the circum fere.net 
of a gnat virde. 



Given a sphere generated by the revolution of the semi- 
circle ACE about the diaiueter AK, with the surface of the 
sphere denoted by S, and its radins by li. 
To prove S==AEX2 nli. 

Proof. Inscribe in the given semicircle the half of a 
regular polygon of an even nnmber of sides, as ABODE. 

Draw the apothem to each side of the semipoiygon, and 
denote it by a. 

From the vertices B, G, D draw Ji to AE. 
Then area AK = AF X2na.\ 

areaflC=F0X2 -^a. 

^ r, ,^j' ■., n 1 Art. 808, 

area CB^0KX2 na. 

&reaDE==KEX2 na. ] 
Adding, area ABGBE^AE X 2 na. 
If, now, the number of sides of the polygon be indefi- 
nitely increased, 

area ABODE approaches S as a limit. Art. m. 
And a approaches E as a limit. (Why f) 

.-. AE X 2 Tta approaches AE X 2 ttK as a limit. (Why !) 
But area ABGDE^AE X 2 jt« always, 

.-. S = AEX2nR. (Why!) 

Q. S. B. 



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SPHEEICA.L AREAS 435 

810. Formulas for area of surface of a sphere^ 
Substitutmg for AE its equal 2 E, S-4 7ti^^ 
Also denoting tlie diameter of the sphere by jD, E = i J). 

:. S=4 7l('~V, orS-TiD^. 



<!)' 



811. Cor. 1. The surface o/g sphere is equivalent to four 
limes the area of a great circle of the sphere. 

812. Gob. 2. The areas of the surfaces of two spheres 
are to each other as the squares of their radii, or of their 
diameters. 

For, if ,S and S' denote the surfaces, E and B' the radii, 
and D and ly the diameters of two spheres, 

8 ^ 4 ^E^ ^ir- 8 ■htC- ^jy^ 

S' inR''^ B'2' ^'^^6" 7ii>'= D'-' 

813. Property of the sphere. The following property of 
the sphere is useil m tlm pi oof of Art. 809: //, in tlie generat- 
ing arc of ami zoni. a biokcn Uiic be inscribed, whose vertices 
divide the arc ndo equal parts, then, as tlie number of these 
parts is increased indefinitely, the area generated by the broken 
line approaches the area of the zone (is a limit. Hence 

Cor. 3. The area of a zone is equal to the circumference 
of a great circle multiplied by the altitude of the zone. 

Thus the area generated by the arc BC = FO X 2 «.B. 

814. Con. -t. On the same sphere, or on equal spheres, the 
areas of Iwo zmn-ti are to each other as the altitudes of the 
zones. 

Ex. 1. Find Ihc iii-c;i nf a sphere wliosi? difimftcr !p 10 in, 

Ex. 2. Fiiid tlif nni\ of a mue of alUtudir ;i io., on a sphere -whose 



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liOOK IX. SOLID 



PKorosiTioM XXV. Theorem 

815. The area of a lime in to the urfa of (he svr~ 
face of ilie sjjhere as the angle of (he hine is to four riijht 




Given a spliere having its area denoted by B, and on the 
epiiere the inne ABCJ) ot £A with its area denoted by L. 
To prove L: S=A° -. 360°. 

Proof. Draw FB3, the great G whose pole is A, inter- 
secting the bounding area of the lime in B and D. 

Case I. When tlte arc BD and the circumference FB3 
are commensurable. 

Find a common measure of BD and FBH, and let it be 
contained in the arc BD m times, and in the circumference 
FBHn times. 

Then arc BD : eirenmference FBII~m -. n. 
Through the diameter AC, and the points of division of 
the circumference FBH pass planes of great © . 

The arcs of these great ® will divide the surface of the 
sphere in » small equal luues, m of them being contained 
in the lune ABCD. 

:. L:8=m:n. 
:. L : jS=arc BD : circumference FBH. (Why ?) 

Or i ; «=.4° : 3G0°. Art. 257. 

Case II, When the arc BD and the circumference FBH 
are incommemurahle . 

Let the pupil supply the proof. q. e. ». 



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SPHERICAL AREAS 4 57 

816. Fonnulafor the area of a lune in spherical degiees, 
or sphends. The sucfaee of a sphere contains 720 Hpheriils 
(Art. 803). Hence, by Art. 815, 
L L spherids 
"720 spherids 

that is, IJie area of a hine in 32>]ier!cal degrees is equal (0 
twice the number of angular degrees in the angle of the liute. 



817. Formula for area of a lune io square units of area. 
■4 7it{""-^m' 



r„ or ;.= 



818. Cor. 1. Oh ihe same Kphere, or on equal .''ijheres, 
two htties (ire to mch other as their anijlex. 

819. Cob. 2. Two bmes with equal angles, but, on mi- 

equal spheres, are to each other as the squares of the radii of 

their spheres. 

'^F-A TtV'^A 
For L : i'.~^ : -^. or /, : L' = B- , E-. 



Ex. 1. Find the area in spherical degrees of a lime of 27°. 

Ex. 2. Find tlie number of square ioches in the urea of a lune oE 
27", on a spliere whose radius is 10 in. 

A solid symmetrical with respect to a plane is a solid in which 
a line drawn from any point in ila surface X the given plane and 
produced its own length ends in a point ou the surface; hence 

Ex. 3. How many planes of symmetry has a circular cylinder f A 
cylinder of revolution ! 

Ez. 4. Haa eillier of these solids a center ot symmetry S 

Ex. 5. Answer the same questions for a circular cone. 

Ex. 6. For a cone of revolution. For a sphere. 

Ek. 7. For a regular square pyramid. For a regular peiitBgonul 



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458 BOOR IS. SOLID GEOMETET 

pROPOsiTiox XXVI, Theorem 

820. If two great circles intersect on a hemisphere, the 
snm of two vertical triangles thus formed is equivalent to a 
bme lohose angle is that angle in the triangles which is 
formed by the intersection of the two great circles. 



Given tbe liemispiiere ADBF, and on it the great circles 
AFB and DFC, intcracetiiig at F. 

To prove A AFC + A BFD ^ luue whose Z is BFV. 

Proof. Complete the sphere aitd produce the given arcs 
of the great circles to intersect at F' on the other hemisphere. 
Then, in the A ^FCand BF'D, 

araAF^RVcBF, 
(each Uing the siipplemenl of the arc BF). 

In like manner arc CF=are DF'. 

And arc AC—ara J)B. 

 : A AFC~ A BF'B. Art. 79a. 

Add the A BFD to each of these equals; 

.-. A AFC + A BFD^A BF'D + A BFD. Ai, 3. 

Or .-. A AF(7+A£FZ)o:luue fBFJ>. Ai. 6, 

Q. £. I). 



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SPHEBICAL AEEAS 459 

Proposition XXVII. Theoreh 

821. The niiinher of spherical dr-grees, or spherids, in 
the area of a spherkal friaugle in equal to (he nnmber of 
angular degrees in the spherical excess of the Irknigle. 



Given the splierical A ABO whose A are denoted by ^, 
B, C, and whose spherical excess is denoted by E. 

To prove area of A ABC =i; spherids. 

Proof. Produce the sides AC and BC to meet AB pro- 
duced in the points D and F, respectively. 
AXBC + A Gl>B = bnie ADDC = 2 A spherids. ) 

> Art. 816. 

AABC+AACF=-hmQ£CI''A := 2 .0Brlieriils. ) 

A ABC + A CFP= luue of Z BOA ^ 2 C spherids. Art. >i2D- 

Adding, and observing that A ABC+A CDJt+AACF 
-I- A (7FD = hemisphere ABDFG, 

2 A ABC+ hemisphere^? {A + B+ C) spherids. Or, 
2 AAS(7+3COsphei-ids=2(.l + 7?+ C) spherids. Ax. s. 

.-. A ABC + 180 spherids^ ( A +h+ C) splierids. Ax. 5. 

.-. A ABC=(A + B+ C—-iiiO) spherids. Ax. 3. 

Ur area A ABC=E spherids. Art. Tss. 



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4(;0 ISOOK 1\. SOLID OKOlIETilV 

822. Formiila for area of a spherical triangle m square 
units of area. 

Comparing the area o£ a spheri<'ul A with the area o£ 

the entire spliei-e , 

area A : 4 %]^ = E splierids : 720 spiierids. 

4 7tR= X 7J . Tin-JE 

:. area A = ^^ , or area A~— rr-r-- 

823. The spherical excess of a spherical polygon is the 
sum of tlie angles of the polygon diminished by (h— 2) 180°; 
that is, it is the sum of the spherical excesses of tiie tri- 
angles into which the polygon may be divided. 



Proposition XXVIII. Theorem 

824. The area of a spherical polygon, in spherical de- 
grees or spherlds, is equal to the spherical excess of the 




Given a spherical polygon AHCDF of n aides, with its 
spherical excess denoted by E. 

To prove area of ABGDF^E spherical degrees. 
Proof. Draw diagonals from A, any vertex of the poly- 
gon, and thus divide the polygon into n—2 spherical A. 
The area of each A= (sum of its i— 180) spherids. 

Art. 821. 

.■. sum of the areas of the A= [sum of /i of the ^ — 

(«— 2) 180] spherids. As. 2. 

.-. area of polygon = E spherids, Art. 823. 

i/or the Diwi I'f A of the A— (.?i— 2) iau°=£°). 



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SPHEIUCAL VOLUMES 



SPHERICAL VOLUMES 

825. A spherical pyramid is a por- 
tion of a spliere bounded by a spheri- 
cal polygon and the planes of the 
great circles forming the sides of the 
polygon. The base of a spherical 
pyramid is the spherical polygon 
bounding it, and the vertex of the 
spherical pyramid is the center of tlie sphere. 

Thus, in the spherical pyramid 0-ABCD, the base i 
ABCD and the vertex is 0. 




820. A spherical wedge (or ungula) is the portion of a 
sphere hounded by a lune and the planes of the sides of 
the lune. 

827. A spherical sector is the portiou of a sphere gene- 
rated by a sector of that semicircle whose rotation generates 
the given sphere. 




828. The base of a spherical sector is the zone gene- 
rated by the revolution of the ai'c of the plane sector which 
generates the spherical sector. 

Let the pnpil draw a spherical sector in which the base 
is a zouc of one buse. 



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4t)l2 T!(K)K IX, WLH> (iEOMETKY 

829. A Spherical segment is a portion of a sphere 
ineluded between two parallel planes. 

The bases of a spherical segment are the sections of the 
sphere made by the parallel planes whlt:h bound the given 
segment; the altitude is the perpendicular distance between 
the bases. 

830. A spherical segment of one base is a spherical seg- 
ment one of whose bounding planes is tangent to the 
sphere. 

Pkoposition' XXTX. Theorem 

831. The voluuu of a -ipheif is equal to oic-ihird the 
product of the area of its suifiice hy its radiU"- 




Given a sphere having its volume denoted by F, sur- 
face by 5, and radius by R. 

To prove 7=iSXK. 

Proof. Let any polyhedron be eireurascribed about the 
sphere. 

Pass a plane through each edge of the polyhedron and 
the center of the sphere. 

These planes will divide the polyhedron into as many 
pyramids as the polyhedron has faces, each pyramid hav- 
ing a face of the polyhedn)H for its base, the center of the 



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SPHERICAL VOLUMES 4R^ 

Sphere for its vertex, siiid the ra<lius of the sphert^ for its 
altitude. 

.'. volume of each pj'ramid = i base X R. (Why ?) 

,'. volume of polyhedron = 4 (surface of polyhedron) XK. 

If the number of faces of the polyhedron be increased 
indefinitely, the volume of the polyhedron approaches the 
volume of the sphere as a limit, and the surface of the 
polyhedron approaches the surface of the sphere as a 
limit. 

Hence the volume of the polyhedron and h (surface of 
the polyhedron) X R, are two variables always equal. 

Hence their limits are equal. 

Or "F-i«Xfi. (Why?) 



832. Formulas for volume of a sphere. Substituting 
^-4 7IE^ or S = 7il>- in the result of Art. 831, 

v= — r— ; also 7=-— -- 



S8S. Cor. 1. The volumes of two spheres are to each 
other as the cubes of their radii, or as the cubes of their 
diameters. 



834. CoK. 2. The volume of a spherical pyramid is 
"/H'tl to one-third the product of its base by the radius of 
thr .yyhere. 

835. COK. 3. Th<' roliioie of spherical ^e'ior is equal to 
■.<„>--lhi,-d the prodKcl of (7.S- haxe (the bomdiiuj zone) by the 
i-<idnis of (he sphere. 



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4G4 



SOLID GEOJIETRY 



836. Formula for the volume of a spherical sector. De- 
noting the altitude of the sector by R ami the volume by T, 

F= J ( area of zone ) X R, 
= J (2 7iRH) n. Art. 8U. 

PROrosiTiox XXX. Theorem 

837. The volume of a spherical segment is equal to one- 
half the product of its altitude by the sum of the areas of 
its bases, plus the coliime of a sphere whose diameter is the 
attitude of tli£ segment. 




Given the semieirele ABCA' which generates a sphere 
by its rotation about the diameter AA'; BD and CF semi- 
chords X AA', and denoted by r and r'; and I>F denoted 
by-ff. 

To prove volume of spherical segmeut generated by 
BCFD, or r=h{7ir^ + 7tr'-) R-{- \ TtiP. 

Proof. Draw the radii OB and OC. 

Denote Of by h, and OD by k. 

Then 7=vol. 0B(7+ vol. OOP— vol. OBB. 

:. y=^ Ttlt-U+^h Ttr'-h—h nr-k. Arts. 830, 723. 

But U=h--li, A^=i^^r'^ andF^B-/--. (Whyf) 



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SPHEIilCAL VOLUMES 465 

= i 7t [2 Sf {li-k)-i (B-S') *— {B"-ri (■]. Ai. a. 
= i Jt [2 E? (;i-f)+ H? (b-D-Ui'-fn- 
= iKj[3ff-(P+M+F)]. 

4'— 2 Ji+F-iP. A-i. 4. 



^3 B-|(,.= +, .»)-?." 



r-j jiH [l ( 



.-. r-i (rtv=+jt.-=)ir+S nH'. 

Q. E. B. 

838. Formula for volume of a spherical segment of one 
base. lu a spherical segment of one base r'~o, and )'"- = 
(2K-ff)ir(Art. 343). 

Substituting for r and r' these values in the result of 

Art. 837, 



'■-"'(^-t)- 



839. Advantage of measurement formulas. The student 
should observe carefully that, by the results obtained in 
Book IX, the iceasurenieut of the areas of certain curved 
sui-faees is reduced to the far simpler work of the measure- 
ment of the lengths of one or more straight lines; in like 
manner the measurement of certain volumes bounded by a 
curved surface is reduced to the simpler work of linear 
measurements. A similar remark applies to the results of 
Book VIII. 

Ex, 1. Find llip volume of a spheve whoso radius is T in. 
Ex. 3. Find tlia voluiHe of a sphere wliose diompt^r is 7 in. 
Ex. 3. In a spliore wIiohb radius is 8 in., find the volume of a 
spheriful segmeiit of one base wliosu aUitiide is 3. 



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466 HOOK TX. fiOT.TD r.F.OMETEY 

EXERCISES. CROUP 72 

THEOREMS CONCRRNlXr. THE SI'lfEliE 

Ex. i. Of circles o£ a sphere whose pLmes pass thronph a civpn. 
point within a sphere, tho smallest is thnt circle whoso plai.s ia 
perpondieAilar to the diameter through tlio given point. 

Ex. 2. If a point on tiie surface of a Riven sphere is equldiRtant 
from three points on a given small circle of the sphere, it is the pole 
of the small circle. 

Ex. 3, It two sides of a spherical triangle are quadrants, the third 
side measures the angie opposite that side in the triangle. 

Ex. 4. If a spherical triangle has one right angle, the sum of its 

other two augles is greater than One right angie. 



Ex. 6. Thfl polar triangle of a biroctangular triangle is hirectan- 
gular. 

Ex, 7. The polar triangle of a trireotangular triangle is ideatieal 
with the original triangle. 

Ek, 8. Prove that the sum of the angles of a spherical quadri- 
lateral is greater than 4 right angles, and less than 8 right angles. 
What, also, are the limits of the sum of the angles of a spherical 
hexagon I Of the sum o! the angles of a spherical n-gon ? 

Ex. 9. On the same sphere, or on equal spheres, two birectan- 
gular triangles are equal if tlieir oblique angles are equal. 



Ex. 1 1. If one of the legs of a right spherical triangle is greate 
than a quadrant, another side is also greater than a qiiaiJr.ant. 

[Sua. Of the leg which is greater than a quadrant, take the em 
remote from the right angle as a pole, and describe an arc] 

Ex. 12. II ABC and A'B'C are polar triangles, the radius OA i 
perpendicular to the plane OB'C. 



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EXERCISES ON THE SPHERE 4b7 

Ex. 13. On tlie same Ephere, or on equal sphecee, spherical tri- 
angles whoae polar triangles have equal perimeters are equivalent. 

Ex. 14. Given OAO', OISO', and All arcs of great 
circles, interseoting so that Z OJC= Z O'BA ; prove that 
^OAB=AO'AI!. 

[SuG. Show that ZOB^I= ZO'JB.] 

Ex. 15. Find the ratio o£ the volume of a sphere to 
the volume oE a circumscribed cuho, 

Ex. 16. Find the ratio of the surface of a sphere to 
the laterr.l surtaee of a ciroumscribod cylinder of revolution; a' 
find the ratio of their volumes. 

Ex. 17. If the edge of a regular tetrahedron is denoted by a, fi 
the ratio of the volumes of the inscribed and circumscribed ephe^j 

Ex. 18. Find the ratio of the two segments into which a hen 
sphere is divided by a plane parallel to the base o£ the UemispliBre b 
at the distance jB from the ba-se. 



EXERCISES. CROUP 73 

Sl'HEKlCAL LOCI 

Ex. 1. Find the locus o£ a point at a given distanne n from the 
Bnrface of a given sphere. 

Ex. 2. Find the locus of a point on the surface of a sphere that is 
equidistant from two given points on the surface. 

Ex. 3. If, through a given point outside a given sphere, tangent 
planes to the sphere are passed, Snd the locus of the points ot 
tangencj. 

Ex. 4. If straight lines be passed through a given Ssod point in 
space, and through anotiier given point other straight lines be passed 
perpecdienlar to the first set, ilud the locus of the Xeet oi the 
perpendiouiuia. 



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4t)0 BOOK IX. SOLID GEOMETRY 

EXERCISES. CROUP 74 

rii015LT:MS COXCl':KMXfi THE STHEliB 
El. 1. At a given point on a sphere, eonatruet a piano tangent to 

the sphere. 

Ex. 2. Through a given point on the siirfnco of a epliere, draw an 

uro of s, great oirele perpendioulac to a givsn are. 

Ex.3. Insoribe a circle in a given spherienl triangle. 

Ex. 4. Construct a spherical triangle, given its polar triangle. 

Given the radius, r, construct a spherical surface which shall pagg 
through 
Ex. 5. Three given points. 



Ex. 6. Two given points and he tangent to a given plane. 

Ex. 7. Two given points and be tangent to a given sphere 

Ex. 8. One given point and be tangent to two given planes. 

Ex. 9. One given point and he tangeut to two given spheres. 

Given the radius, r, constrni^t a spherical siirfacQ which shall be 
tangent to 
Ex. 10. Three given planes, 
Ex. 11. Two given planes and <jne given sphere. 

Ex. 12. Construct a spherical surface whiah shall pass through 
three given points and be tangent to a given plane. 

Ex. 13. Through a given straight line pass a plane tangent to a 
given sphere. 

[Suo. Through the center of the sphere pass a plane J. [tiven 

When is the solution iinpossiiule ? 

Ex. 14. Through a given point on a sphere, construct an aru of 
a great oirele tangent to a given small circle of the sphere. 

[Sue, Draw a straight line from the center of the sphere to the 
given point, and produce it to intevseot the plane of the small oirele, 
etc.] 



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NUMERICAL EXERCISES IN SOLID 
GEOMETRY 

For methods of facilitating immerical computations, 
see Arts. 4i)3-6. 

eXEROISES. CROUP' 7S 

LINES AND SURFACES OP POLYHEDRON'S 
Find the lateral area and total area oE a riglit prism whose 

Ex. 1. Ease is au. equliatoral triaiigly of edge 4 in., aud whose 
altitude is 15 in. 

Es. 2, Base is a triangle of aides 17, 12, 25, aud whose altitude 
is 20, 

Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10 
and 15 and leg 8, and whose altitude is 24. 

Ex. 4. Base is a rhomhus whose diagonals are 12 and 16, and 
■whose altitude is 12. 

Ex. 5. Base is a regular hexagon wUh side S tt., ar.i whose alti- 
tude is 20 ft. 

Ex. 6. Find the entire surface of a reetanguiar parallelepiped 
8Xl2XlGiii.; ofonepX'jXrft. 

Ex, 7. Of a eube whose edge is 1 ft, 3 in. 

Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft. 
and an ed^-e o£ the base is 10 tt. Find the altitude. 

Ex. 9. How many square feet of tin are neeessary to line a box 
20X6X4in.? 

Ex. 10. 1( the surfac'e of a c^ube is 1 pq.vd,, find an edge in inehes 

E!t, 11. Find the diagonal of a cuhti whoso edge is 5 in. 

Ex. 12. I£ the diagonal of a cube is 12 ft., find the surface. 



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470 HOLID GEOMF.TliY 

Ex. 13. If tho surfaop oEnreowngiilaipnrallolopipedia 208aq. in., 
and the odg03 are aa 2 ; y : 4, liud tUu oJjskh. 

In a regular square pyramid 

Es. 14. IE an udge o£ the base ia Id and eUut height is 17, find 
the ftltitude. 

nteral edge is 17, find nn edge 

Ex. 16. If i\ lateral edge is 2:i aud an edge of the Ijiae is 14, find 
the altitude, 

In a regular ti'iangular pyramid 

Ex. 17. If an edge of the baso is 3 and the altitude ia 10, find tho 
slant height. 

Ex. 18. Find the altitude of a regulav tetrabudron whouo edge \i 6, 
Find the lateral surface and the total surface o£ 

Eic. 19. A regular square pyramid an edga ot wlioso base is 16, 
and whose altitude is 15. 



Ex. 22. A regular square pyramid whose slant height is 24, and 
ihose laterai edge is 25. 
Ez. 23. A regular tetrahedron whose edge ia 4. 
Ex. 24, A regular tetrahedron whose altitude ia 9, 



Ex. 26. In the frustum of a regular square pyramid the edges of 
the bases are 6 and 18, and the altitude ia 8. Find the slant height. 
Hence find the lateral area. 

Ei, 27. In tho fru.^tum of a regular triangular pyramid the edges 
ot the bases are i and 6, and the altitud. ia 5, Find the slant height. 
Hence find the lattirul a-ttat. 



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NUMERICAL EXERCISER IN SOLID GEOMETRY 471 

Ex. 28. In tlie frustum of a regular tetrahedron, if the edge of the 
lower base is hi, the edge of the upper base is 63, and the altitude is a, 
show that L = il/i{b, — b-i)' +ia'. 

Ex. 29. lu the fruatum of a regular square pyramid the edges o£ 
the bases are 2U and 60, and a lateral edge ia 101, Find the lateral 



EXERCISES. CROUP ?e 

LINES AND SURFACES OP CONES AND CYLINDERS 

Ex. 1. How many square feet of lateral surface has a tunnel 100 
I. long and 7 ft. in diaiueteiv 



In a cylinder of revolution 

Ex. 4. Find E in terms of S and H. 

Es. 6. Find H in terms of E and T. 

Ex. 6. Find T in terms of 5 and il. 

Ex. 7. How many sq, yds of canvas are required to make a coni- 
oal tent 20 ft. in diameter and 12 ft. liigb ! 

El. 8. A man has 400 aq. yds. of canvas and wants to make a 
conical tent 20 yds. In diameter. What will be its altitude 1 

Ex. 9. The altitude of a cone of revolution ia 10 ft. and the lat- 
eral area is 11 times the area of the base. Find the radius of the baae. 
In a cone of reTOlution 

Ex. 10. Find T In terms of S and L. 

Ex. 1 1. Find E in terms of T and L. 

Ex. 12. How many square feet c 
funnel the diametere of whoso ends 
altitude ia 7 in. ? 

Ex. 13. If the slant height of a fnistnm of a eoue of  
makes an angle of 45" with the base, show that the lateral a 



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SOLID GEOMETRY 

EXERCISES, CROUP ^T 

RPHr;R!CAT- LINES AND SL'RrACi:^ 



El, 2, How many squaro inelies of leather will It take to eovar a 
bafieball whose diameter is 3^ in.f 

Ex. 3. How many aq. ft. of tin are required to covpr a dome in 
the shape of a liemiaphere 6 yds. in diameter ? 

Ex. 4. What ia the radius of a sphere whose surface is filG sq. in. f 

Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd. 



Ei. 7. If a hamispherical dome is to contain 100 sq. yds. of sur- 
face, what must its diameter be V 

Ex. S. Find the radius of a sphere in which the area of (he sur- 
face equals the number o£ linear units in the circumference of a great 

Find the area of a !une in which 

Ex, 9. The angle of the lune is 3(i°, and the radius of the sphere 
is 14 in. 

Ex. 10. Tlie angle of the lune is lfi° 'M, and the diameter of the 

Ex. 11. The angle of the lune is 24°, and the surface of the 
sphere is i sq. ft. 

Find the area of a spherical triangle in which 

Ex. 12, Tiie angles are 80^, 90", 120°, and the diameter of the 
sphere ia 14 ft. 

Es. 13. The angles are 74° 24', 83° 16', 92° 20', and the radius of 
the sphere is 10. 

Ei. 14. The angles are 85°, S5°, 135°, and the surface of tlie 
epliere ia 10 aq. ft. 



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NUMERICAL EXERCISES IN SOLID GEOMETEY 473 

Ex. 15. If the sides of n spherical trianKle nre 100°, 110^, 120" 
and the radius of the sphero U 16, find the area of the polar triangle, 

Ex. 16. It the angles of a spherical triangle are 00", 100". 1L>0'* 
and its area is Bi)00, fiud tho radius of the sphere. 

Ex. 17. If the area of an equilateral apherioai triangle is one- 
third the surface of the spliere, fled an angle of the triangle. 

Ex. 18. In a trihedral angle the plane angles ot the dihedral 
angles are 80", i>0°, 100°; find the number of solid degrees in the 
trihedral angle, 

Ex. 19. Find the area of a spherical hexagon eanh of whoaa 

Ex. 20. If each dihedral angle of a given pentahedral angle is 
120", how many solid degrees does the pentahedral angle contain 1 

Ex. 21. In a sphere whose radius is 14 in,, fiud the area of a zono 
3 in, high. 

Ex. 22. What is the area of the north temperate zono, if the 
earth is taken to be a sphere vrith a radius of 4,000 miles, and tlie 
distance between the plane ot the arctic circle and that of the tropic 
of Cancer is 1,800 miles t 

Ex, 23. If Cairo, Egypt, is in latitude 30°, show that its parallel 
ot latitudft bisects the surface of the northern hemisphere. 



Ex. 25. How much of the earth's surface will a man see who is 
,000 miles above the surface, if the diameter is taken as 8,000 milesS 

* of a zone equals the area of a groat circle, 
zone in terms of the radius of the sphere, 

Ex. 27. If sounds from the Krakato a explosion were heard at a 
distance of 3,000 miles (taken as a chord) on the surface of the earth, 
over what fraction of the earth's surface were they heard 1 

Ex. 28. The radii of two spheres are 5 and 12 in. and their cen- 
ters are 13 in, apart. Find the area of the circle of intersection and 
also of that part of the surface of each sphere not Included by the 
otiier ephere. 



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474 SOLID GEOMKTKY 

EXERCISES. CROUE' 7a 

VOLUMES OF POI.YHEDKONS 
Find the volume of a priam 

Ex. 1. Whose base is an equilateral triangle ivitli Hide 5 !n., and 

1j, and whose 

Ex. 3. Whose bnae is an isoaeelea rigbt triangle with a leg equal 
to 2 yds., and whose altitude is 25 ft. 

Ex. 4. Whose base is a regular hexagon wvitli a side of 8 ft , and 
whose altitude is 10 yds. 

Ex. 5. Whose base is a rhombus one of whose sides is 2ii, and one 
of whose diagonals is 14, and whose altitude is 11. 

Ex. 6. Whose base contains 84 bc;. yds., and whose lateral fanes 
are three rectangles with areas of 100, 170, 210 sq. yds., respectively. 

Ex. 7. How many bushels of wheat are held by a bin 30x10x6 ft., 
if a tii^hel is taken as li cu. ft.? 

Ex. 8. How many eart-loads of earth are in a collar 30 s 20 x G ft, 
if a uart-load is a cubic yard ? 

Ex. 9. If a etibioftl block of marble coats $3, what is the cost of 
a tube whose edge is a diagonal of the first blocl; t 



Ex. II. Find the edge of a cube whose volume equals the areii of 
its surface. 

Ex. 12. If the top of a cistern is a rectangle 12 s 8 £t., how deep 
must the cistern be to hold 10,000 gallons F 

Ex. 13. Find the inner edge of a peck measure which is in the 
shape of a cube. 

Ex. 14. A peck Taeasure is to be a rectangular patallelopipod with 
square base and altitude equal to twioe the edge of the base. Find 



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NUMEKICAL EXEHCISES IN SOLID GEOMETRY 475 

Ex. IS. Find the volume of a cube whose diagonal is a. 
Find the rolurae of a pyramid 

Ex. 16. Whose base is an equilateral triangle with side 8 in,, 
and whose altitude is 13 in. 



Ex. 17. 
one leg 21, 


Whose 
and wh 


1 base is a right 1 
ose altitude ie 20, 


:rianele 


with 


hypotei 


™. 


>e 29 an 


,d 


Efc. 18. Whose base 
lateral edges is 5. 


is a square 


with i 


iide (>, 


and ea. 


?h 


of who' 


.e 


Ex. 19. Whose 
lateral fa^es mal^e 


. hase 


is a square 
ngle of 45° V 


with siiJe 10, 
tith the base. 


and en 


ith 


of whoi 


,e 


Ex 20 


H the pvca' 
re b^ise of si. 
tain r What 


mid of Memphis ha 
ae 2:i;!yd3., how m 
is this worth at $1 b 


IS an altitude 
■any cubic ya 
>cu. yd.? 


of 

rds 


146 yd 
of stoi 





Ex. 21. A ehiirch spire 150 ft. high is hesagonal in shape and 
each side of the base is 10 ft. The spire has a hollow hexagonal 
interior, eanh side of whose base is 6 ft., and whose altitude is 45 ft. 
How many oubic yards of stone does the spire contain 1 

I. yds. and its base is a square 



Ex. 23. A heap of candy in the shape of a frustum of a regular 
square pyramid has the edges of its bases 2ri and !] in . and its altitude 
12 in. Find the number of pounds in the heap if a pound is a rectan- 
gular parallelepiped 4x3x2 in. in size. 

Ex. 24. Find the yoluma of a frustum of a regular triangular 
pyramid, the edges of ths bases being 2 and 8, and the slant height 12. 

Ex. 25. The edgea of the bases of the frustum of a regular square 
pyramid are 24 and G, and each lateral edge is 13 ; Snd the volume. 

Ex. 26. If a stick of timber is in the shape of a frustum of a 
regular square pyramid with the edges of its ends and 15 in., and 
with a length of 14 ft., End the number of feet of lumber in the stick. 

What is the difference between this volume and that of a stick of 
the sumo length having the shape of a prism with a base equal to tie 
area o! a midsection ot the first sUck ; 



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Ex.. 28. How many uavt-loada o£ eartli are there in n railroad cut 
12 ft. deep, n-liose base is a rectangle 100 x 8 £t., and whose top is a 
reetangle 30 x 50 It, I 

Ex. 29. Find the volume ot a prismatoid whoso baae is an equi- 
lateral triangle with aide 12 ft., nnd whose top is a Una 12 ft. long 
liarallel to one Bide of the base, and whose altitude is 15 ft. 

Ex, 30. If the haae of a prismatoid is a rectangle with dimensions 
a and b, the top is a line c parallel to the aide !i of the liase, and the 
altitude ia 'i, find the vulume. 

EXERCISES. CROUP ta 

VOLTTJIES OF COXES AND CYLlNDii;RS 

Ex. 1. How many barrels of oil are contained in a cjlicdrical 
tank 20 ft. long and G ft. in diameter, if a barrel contains 4 cu. ft. t 

Ex. 2. How many ou. yds. of earth must be removed in making a 
tunnel 450 ft. long, it a eross-seotion of the tunnel is a aeiEiicirole of 
15 ft. radius f 

Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint. 
Find its height in inches. 



Ex. 4. If a cubic foot of 
lameter, how long will the w 



altitude equals the 



Ex. 6. Showthat the YOlumesof two cylinders, having the lUtitude 
f each equal to the radius of the other, are to each other as E : S'. 



Ex. 8. A conical heap of potatoes is 44 ft. in circumfeienee and 
6 ft. high. How many bushels does it contain, if a bushel is lieu. It.t 



e- glass hold, it 



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NTTMEEICAL EXERCISES IN SOLID GEOMETItY 477 

Ex. 10. Find the ratio of tho volumea o{ the two eonts ins^iribed 
ribed about, a regular tetrahedron. 



X. 11. IE an equilateral eono eontaius 1 quart, find its diraen- 

b of R aud L ; alao 

Ejc. 13. Find tiie volume of a frustum of a cone of revolution, 
whose radii are 14 and T ft., and whose altitude is 3 yds. 

Ex. 14. What is the cost, at 50 cts. a oh, ft., of a piece of marble 
in the shape of a fruBtum of ft cone of revolution, whose radii are 6 
and 9 ft., and whose slant height is 5 ft.? 



ExenciSEs. crouf so 

SPHERICAL VOLUJLES 
Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in. 
Ex. 2. I! the earth is a sphere 7,9L'0 miles In diameter, Bnd iti 

Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft. 

Ex. 4. What is the volume of a sphere whose surface isClG sq. in.' 

Ex. 5. Find the radius of a sphere equivalent to the sum of twi 
pheres, whose radii are 2 and 4 in. 



Ex. 7, Find the volume of a sphere oircumseribed 
whose edge is fi. 

Ex. 8. Find the voluiue of a spheiifial shell whc 
outer diameters are 14 and 21 iu. 

Ex. 9. Find the volume o( u spberii.'al shell whoso ii 
surfaces :izo 20 tt wnd i: tt. 



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4 1 8 SOLID GKOJIETllY 

Ifind tiio voluBid of 

Es. 10. A splierieal wedge whoso auglo is 3!", the radius of the 
sphere being 10 in. 

Ex. 11. A spherical Keetor whoso base is a. zone 2 in. high, the 
rartius of the sphere being 10 in. 

a whose rsdii are 4 and 



Ex. 13. A wash-basin in the shape r>£ a aogment, of a sphere is 
6 in. deep and 2i in. in djonieter. How many quarts o£ water will the 
basin hold t 

Ex. 14. A plana parallel to the base of a hemisphere and bisect- 
ing thi! altitude diifides its volume in what ratio f 

Ex. 15. A aplieri<^al Kogmeut 4 in. high «oiitains 200 eu. in. ; End 
the r.idiu.'i of the sphere. 

Ex.16. If a heavy sphere whoso dia 
ponieal wice-Glass full of water, whose i 
6 in., find how much water will run over 



EXERCISES. GROUP 81 

EQUIVALENT SOLIDS 



Ex. i. IE a cubical block of putty, eaffh edge ot whi 


ch is 8 incheg, 


be molded into a cylinder oE revolution whose radius h 


1 3 inches, find 


the altitude ot the eyiicder. 




Ex. 2. Find the radius of a sphere equivalent to 


a cube whose 


edge ie 10 in. 




Ex. S. Find the radius of a sphere equivalent to a i 


;one of revolu- 


tion, whose radius ia 3 in. and altitude G in. 





Ex. 4. Find the edge of a cube equivalent to a frustum of a eons 
f revolution, whose radii are 4 and 9 ft. and altitude 2 yds. 

Ex. 5. Find the altitude of a rectangular parallelopiped, whose 
ase is 3x3 in. and whose volume Is eqaivaleot to a spheru of tadiua 



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NUMERICAL EXERCISES IN SOLID GEOMETlil' 479 

Ex. 6. Find the biiRe of a square reetangular pai'allelopiped, 
Vfliose altitude ia 8 in. aud whose volume equals the volume o£ a cone 
of revolution with a radiuB of 6 and au aititudo of 12 m. 

Ex. 7. rind the radius of a cone of revolution, whose altitude is 
15 and whose volume is equal to that of a eyliniier of revolutiou ivith 
radius G and altitude 20. 

Ex. 8. Find the altitude of a eone of revolution, whoso radius is 
15 and whose voiume equals the volume of n cone of revolulion with 
radius 9 and altitude 24. 

Ex, 9. On a sphere whose diameter is 14 the altitude of a zone of 
one base is 2. Find the altitude of a pylinder of revolution, whose 
base equals the base of the zone and whose lateral surface equals the 
surface of the zone. 



EXERCISES. CROUF* 

SIMILAE SOLIDS 



Ex. 1. If on two similar solids L, L' and !, I' are 
ORous lines; -4, A' and a, «' pairs of homologous 
T, v' pairs of homologous volumes, 


pairs of homol- 
veas, V, V and 


ft :L' = 1 :I' = v'-^ ■.i/A' = ^'v.f'e'. 
show that - A : A' = L^ : L" = a : n'= r* : FL 




Ex, 2. If the edge of a cube is :o in., &nd the 
having 5 times the surface. 


edge of a cube 



Ex. 3, If the radius of a sphere is 1(1 in., find the radiu 
Bphere having 5 times the surface. 



Ex. 5. In the last three exercises, find the required di 
10 volume ia to be 3 times the volume of the original solid. 



those of auotliui- trunk. How 



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SOTJD GEOMliTUy 



Ex. 7. How fill 


 from Iho vertex is 


i Ihe croBK-section which bisects 


tho volume of a 


cone Ot I'BVOlutiOI 


1 f Wliich biseeta tiie lateral 


Eurfaee ? 







Ex. 8. If the nltiliide oC a pyramid is bisected Tjy a plane parallel 
to the 1)sse, how does the area of tho cross-seetion ootupare with the 
area of the liase ? How does the volume cut olf compare with the 
volume of the entire pyrnmid ? 

Ex. 9. Plnnes parallel to Iho base of a coae divide tho altitude 
into three equal pai'ts ; compare the lateral surfaces cut off. Also the 
volumes'. 

Ex. 10. A sphere 10 i(i. in diameter ia divided into three equiva- 
lent parts by concentric spherical surfaces. Find the diameters of 
toese surfaces. 

Ex. 11. It the strength of a muscle is as the area of its cross- 
section, and Goliath of Gath was three times as large ia each linear 
dimension as Tom Thumb, how much gre;iter was his strength ! His 
weight f How, then, does the activity of the one man compare with 
that of the other ? 

Ex. 12. If the rate at which heat radiates from a body is in pro- 
portion to the amount of surface, and the planet Jupiter has a diame- 
ter 11 times that of the earth, how many times longer will Jupiter be 
in cooling off t 

[SuG. How many times greater is the volume, and therefore the 
original amount of heat in Jupiter ? How many times greater is its 
surface f What will be the combined effect of these factors T] 



CROUP 83 

MISCELLANEOUS NUMERICAL EXERCISES IN SOLID 
GEOMETRY 
Find S, T and V of 

Ex. 1. A right triangular prism whose altitude is 1 ft., and t! 
Bides of whose base are 26, 28, 30 in. 

base ia 1ft. 2ii 

Bx. 3. A frustum of a square pyramid Iho areas of whose has 
are 1 sq. It, and 3G sq. in,, and whose altitude ia !) ia. 



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NUMERICAL EXEECISES IN SOLID GEOMETRY 48X 

Ex. 4. A pyramid whose slant height is 10 in., and whose base 
I an equilateral triangle whose side is 8 in. 



Ex. 6. A frustum o( a cooe of revolutioo whose radii 
11 in. and slant height 13 in. 



Ex. 8. Find the volume of a sphere inscribed in a cube whose 
edge is 6; also find the area of a triangle on that sphere whose anglea 
are 80°, 90", 150°. 

> of the spherical pyramid whose base is 



Ex. 10. Find the angle of a lime on the same sphere, equivalent 
to that triangle. 

Ex. II. On a cube whose edge is 4, planes through the midpoints 
of the edges eut oE the corners. Find the volume of the solid re- 
maining. 

Ex. 12. How is F ehnnged if n of a cono of revolution is doubled 
and R remains unchanged t If R is doubled and W remains unchangedt 
If both H and J! are doubled ? 

Ex. 13. In an equilateral cone, find S and V in terms of E. 

Ex. 14. A piece of lead 30 x H :t 3 in. will make how many spher- 
ical bullets, each S in. in diameter ! 

Ex, 16. How many bricks are necessary to make a chimney in 
the shape of a frustum of a eone, whose altitude is DO ft., whose outer 
diameters are 3 and 8 ft,, and whose inner diameters are 2 and 4 ft., 
eonnting 12 bricks to the eubio ft. ? 



a of a zone is 300 and its altitude 6, find t 



Ex. 18. If every edge of a square p7iamidisit Qtid^in termsof T. 



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453 SOLID GKOMliri'llI: 

Ex, 19, Arpgiikr ^qHarc pyramid hns a for its allituJe and also t ,,. 
encli side of the liiu-e. Find llie area of a section made by a plane parallel 
to the base mid bifieGting Ilia altitude. Find also tlie volumea of tha 
two parts into wliieli the pyramid ia divided. 

Ex. 20. It the earth is a sphere of 8,000 milos diameter and its 
alroosphere extends 50 miles from the eavlb, find the volume of the 

Ex. 21. On a spherp, find the ratio of llie area or an e.iuilateral 
ipberipal Iriiinglp. eaeh of wliose angles ia 35', to liie area of a lune 

Ex. 22. A square right prism has an altitude Cm and an edge o£ 
the base 2a. Find the volume of the largest cylinder, sphere, pyramid 
ami cone which can be ent from it. 

Ex. 23. Obtain a formula for the area of that part of a'sphere 
illujuinateit by a jioint o£ light at a distance a from the sphere 
whose radins is H. 

Ex. 24. On a sphere whose radius is 6 in., find an angle of an 
equilateral tnaBjtle whose area is 12 sq. in. 

Ex. 25. Find the volume of a priemiitoid, whose altitude is 2-1 and 
whose bases are equilateral triangles, each side 10, so placed that tlie 
raid-seotion of the prismatoid is a regular hexagon. 

Ex. 26. Ou a sphere tihose radius is 16, the bases of a zone are 
equal and are together equal to the area of the zone. Find the alti- 
tude of the zone. 

e of a square pyramid, the edge of whose 
teral edges is inclined G0° to the base. 

Ex. 28. Aq irregular piece of ore, if placed in a cylinder partly 
filled with water, causes the water to rise 6 in. If the radius of the 
cylinder is 8 in., what is the volume of the ore ? 

Ex. 29. Find the volume of a truneated right triangular prism, if 
the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14. 

Ex. 30. In a sphere whose radius is 5, a aectioii is taken si tho 
distance 3 from the center. On this section as a base a oone iu formoti 
whose lateral elements are tangent to the sphere. Find tho lateral 
surface and volume of the coua. 



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NUMERICAL EXEECIRES IS SOLID GEOatETRY 483 

Ex.31. The volume of a sphere 13 l,437iau. in. Find the surface. 

Es. 32. A square whose side is 6 is revolved about a diagonal as 
a axis ; find the surface and volume generated. 

Ex. 33. Find the edge ot a cubical cistern that will hold 10 tons 
f water, if 1 cu. ft. of water weighs 62.28 lbs, 

Ex. 34. A water trough has equilateral triangles, eaeh side 3 ft., 
)r ends, and is IS ft. loiip. How many buckets of water will it hold, 
! a bucket ta a cylinder 1 ft. in diameter and li ft. high J 



Ex. 35. Tlie lateral ar. 


ea of a (.-yiinder of r 


evolution is 440 sq. 


d the volume is 1,5411 cu 


.. in. Find the ra.lli 


IS and altitude. 



Ex. 36. The anfrles of a spherical quadrilateral are 80", 100", 
1-0", 120°. Find tile angle ot an equivalent equilateral triangle. 

Ex.37. A cone and a cylinder have equal lateral suifacep, and 
their axis sections are equilateral. Find the ratio of their volumes. 

Ex.38. A water-pipe I in. 1 
How many quarts of water mu 
from under the ground oemes i 
how long must the watei' run f 

Ex. 39. A cube immersed 
causes the water to rise 4 in. 
what is an edge ot the cube 1 

Ex. 40. An auger hole whos 



Ex. 42. The volumes ot two similar cylinders of revolution are 
as 8 : lliS; find the ratio of their radii. If the r.tdius of the smaller if 
10 in., what is the radius of the larger? 

Ex. 43. An iron shell is 3 in. thick and the di^imeter of Us oute: 
surface is 28 in. Find its volume. 

Ex. 44. The legs of an isosceles spherical triangle each make fir 
".ngle of Ta° «ith the base. The legs [>i'odueed form a iiine whosi 
aiea is four times the area of the triangle. Find the angle of the lune 



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484 SOLID CEO^iirrsY 

CROUP 81. 

EXERCISES INVOLVING THE MISTRIC S^YSTEM 

rind s. r, r o( 

Ex. 1. A riKht priam tho edRes of nhoso bas^o are li m., 70 dm., 
900 cm., and wbose altitude is 90 dm. 

Ex. 2. A regular square pyramid sn fidge of whose base is 30 lim., 
and whose altitude is 1.7 m, 

Ex. 3. A sphere whose radius is 0.02 m. 

Ex. 4. A frustum of a oone of reTolution whose radii are 10 dm 
acd 6 dm., and whose slant height ia 50 cm. 

Es. 5. A cube whose diagonal is 12 cm. 

Ex. 6. A cylinder of revolution whose radius equals 2 dm., auii 
■whose altitude equals the diameter of the base. 

Ex. 7. Find the area of a sphericnl trianglp on a sphere who^f 
radius is 0.02 m., if its angles are 110°, 120°, 130°. 

Ex. 8. Find the number of square meters iu the surface of [ 
sphere, a great circle of which is 50 dm. long. 

Ex. 9. How many liters will a cylindrical vessel hold that i 
10 dm. in diameter and 0.iJ5 m. high ? How many liquid quarts ? 

a cylinder whose diameter is half th. 
Ex. 1 1. Find the surface of a sphere whose TOlume is 1 cu. m. 



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APPET^DIX 

I. MODERN GEOMETRIC CONCEPTS 

840. Modern Geometry, In recent times many new 
georaetfie ideas have been invented, and some of them 
developed into important new branches of geometry. 
Thus, the idea of symmetry (see Art. 484, etc.) is a 
modern geometric concept. A few other of these modero 
concepts and methods will he briefly mentioned, but 
their thorough consideration lies beyond the seope of 
this hook. 

841. Projective Geometry. The idea of projections 
(see Art. 345) has been developed in eomparatiyely 
recent times into an important branch of mathematics with 
many praetieal applications, as in engineering, architec- 
ture, constructicm of maps, etc. 

842. Principle of Continuity, By this principle two 
or more theorems are made special cases of a single more 
general theorem. An important aid in obtaining continuity 
among geometric principles is the application of the con- 
cept of negative quantity to geometric magnitudes. 

Thus, a negative line is a line opposite in direction to 
a given line taken as positive. 



For example, if OA is +, OB is - 
(485) 



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48G 




 Similarly, a negative angle is 
an angle formed by rotating a line 
in a plane in a direction opposite 
from a direction of rotation taken 
as positive. Thus, if the line OA 
rotating from the position OA 
forms the positive angle AOB, the 

same line rotating in the opposite direction forms the 
negative angle AOB'. Similarly, positive and negative 
ares are formed. 

In like manner, if P and P' are 
on opposite sides of the line AB and 
the area PAB is taken as positive, 
the area P'AB will be negative. 

As an ilhistratiou of the law of 
continnity, we may take the Uieoreni 
that the sum of the triangles formed 
by drawing lines from a point to the 
vertices of a polygon equals the area 
of the polygon, " " 

Applying this to the qnadrilateral ABCD, if the point 
Pfalla within the quadrilateral, ^PAB+AFBC + APCI) 
+ i^rAD=ABCr> (Ax. 6). 

Also, if the point falls without the quadrilateral at J", 
AP'AB + AP'BG+ AP'C7>+ A P'AD == ABCD, since 
APAD is a negative area, and hence is to be subtracted 
from the sum of the other three triangles. 

843. The Principle of Reciprocity, or Duality, is a 
principle of relation between two theorems by which each 
theorem is convertible into the other by causing the woi-ds 
for the same two gooniotrie objects in each tlicorem to ex- 
cliaTige plaecK. 




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MODERN GEOMETRIC CONCEPTS 487 

Thus, of t.liL'orems VI and VII, Book I, either may be 
convei'ted into the otiier by replacing the word "sides" by 
"aogles," and "angles" by "sides." Henee these are 
termed reciprocal theorems. 

The following are other instances of reciproeiil geometric 
properties : 



1. Two points 

2. Three p<ri«U 
straight line deleriN 


determine a 
<e a plane. 


1. Tw 

2. Tin 


lines lie t ermine a point. 

ee plaim 7inl through the 
ight line ileierniine a 


3. A straight U 
AeUnnine a piune. 


" <""' « I'O'"' 


3. Js 
deter iiiiiw 


fraighl line wul a plane- 
a i'oiiit. 



The reciprocal of a theorem is not necessarily true. 

Thns, two parallel straight lines determine a plane, but 
two parallel planes do not determine a line. 

However, by the use of the principle of reciprocity, 
geometrical properties, not otherwise obvious, are fre- 
quently suggested. 

844. Principle of Homology. Just as thii law of reci- 
procity indicates relations between one set of geometric 
concepts (as lines) and another set of geometric concepts 
(as points), so the law of homology indicates relations 
between a set of geometric concepts and a set of concepts 
outside of geometry: as a set of algebraic concepts, for 
instance. 

Thus, if a and l are nnmbers, by algebra {a + h) (a — i) 

Also, if a and b are segments of a line, the rectangle 
(«+6)X (i( — h) is equivalent to the difference between the 
squares ir and V'. 

By means of this principle, truths which would be over- 
looked or difficult to prove in one departmeut of thonght 



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488 (lEO^liriKY. Al'I'ENVlX 

are made obvious by observing tlie corresponding truth in 
another department of thought. 

Thus, if a and h ai-e line segments, tlie theorem (a + h)" 
+ {a—b}- — 2{(t--i-b-} is not immediately obvious in geo- 
metry, but becomes so bs observing the Hke relation 
between the algebraic uumbers a and i>. 

845. Non-Euclidean Geometry. Hyperspace. By vary- 
ing the properties of space, as these are ordiuarily stated, 
different kinds of space may be conceived of, each having 
its own geometric laws and properties, Thus, space, as 
we ordinarily conceive it, has three dimensions, but it is 
possible to conceive of space as having four or more 
dimensions. To mention a single property of four dimen- 
sional space, in sneh a space it would be possible, by 
simple pressure, to turn a sphere, as an orange, inside 
out without breaking its surface. 

As an aid toward eonoeiving how tliia ia poaslblo, consider a plane 
in wiiicil one eii'cle lies inside another. No matter how these eirelea 
are moved about in the plane, it is impossiljle to shift the inner eirale 
KO aa to piaee it outside the other witliout l)rea!iing the oiroumference 
III the outer circle. But, if we are allowed to use the third dimension 
of space, it ia a aimple matter to lift the inner circle up out of the 
plane and set it down outside the larger oirale.' 

Suuilarly it, in space o£ three dimensions, we have one spherical 
aliell inside a larger shell, it is impossible to place the smaller shell 
ontsida the larger without breaking the larger. But if the nse of a 
fourth dimension be allowed, — that is. the use of another dimension 
of freedom of motion, — it ls possible to place Che inner shell outside 
the larger without breaking the latter. 

846. Curved Spaces. By varying the geometric, axioms 
of space (see Art. 47), different kinds of space may be 
conceived of. Thus, we may conceive of space such that 
through a given point one line may be drawn parallel to a 
given ime (that is ordmary, or Euclideau space}; or such 



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MODF.KS GEOMETillC COKCEPTS 489 

that through a given point no line can be drawn parallel 
to a given liue (spherical space); or such that through a 
given point more than one line can be drawn parallel to a 
given liue (pseudo- spherical space). 

These different kinds of space differ in many of their 
properties. For example, in the first of them the sum of 
the angles of a triangle equals two right angles; in the 
second, it is greater; in the third, it is less. 

These different kinds of space, however, have many 
properties in common. Thus, in all of them every point in 
the perpendieaiar hiseetor of a liue is equidistant from the 
extremities of the liue. 

eXERCISKS. CROUP SE 

Ex. 1. Show by the use of zero and negative ares that the princi- 
ples of Arts. 257, 263, 258, 264, 203, are particular cases of the genei'al 
theorem that the angle inoluded between two lines which cut or tooth 
a cirule is measured hj oue-lialf the sum of the intercepted area. 

Ex. 2, Show that the principles of Arts. 354 and 3o8 are particular 
eases of the theorem that, if two lines are drawn from or through a 
po'nt t m t a eireumfa ranee, the product o£ the segments o£ one line 
equ h p duct of the segments of the other iino. 

Ex 3 bhow by the use o£ negative angles 
tha h em XXXVIII, Book I, is true for a 
quad a e al o£ the form AISCD. IBGD is a 
ne a e an le; the angle at the vertex i» is 
the fl sa leADC] 

Ex 4 What is the reciprocal of tiie state- 



Ex. 5, What is 
perpendicular to i 
dicular to eiich oti 




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IT. ITIRTORY OP GKO:\IETRY 

847. Origia of Geometry as a Science. The beginumgR 
of geometty as a science are found in Kgypt, diitiug back 
fit least tliree thonsaud years before Christ. Herodotus 
says that geometry, as known iu Egypt, grew out of the 
need of remeasuring pieces of laud parts of which had been 
washed away by the Xiie floods, in order to make an e^ni- 
tabie readjustment of the taxes on the same. 

The substance of the Egyptian geometry is found in an 
old papyrus roll, now iu the British museum. This roll 
is, in eSect, a mathematical treatise written by a scribe 
named Ahmes at least 1700 B.C., and is, the writer states, 
a copy of a more ancient work, dating, say, 3000 B. C. 

848. Epochs In the Development of Geometry. Prom 
Egypt a knowledge of geometry was transferred to Greece, 
whence it spread to other countries. Hence we have the 
following principal epochs in the development of geometry: 

1. Egyptian : 3000 B. C— 1500 D. C. 

2. Greek : COO B. C— 100 B, C. 

3. Hindoo : 500 A. D— 1100 A. D. 

4. Arab : 800 A. D.— 1200 A, D. 

5. European : 1200 A. D, 

In the year 1120 A. D.. Athelard, an English monk, 
visited Cordova, m Spam, in the disguise of a Mohamme- 
dan student, and proeui-cd a copy of Euclid m the Arabic 
language. This book he brought back to central Europe, 
where it was translated into Latin and became the basis of 
all geometric study in Europe till the year lriJ3, when, 
(WO) 



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niKTORY OF GEOMETRY 4i'l 

owing to the capture of Ooiiatautiiiople by the Turks, 
copies of the worlis of the Greek mathematicians in the 
original Greeli were scattered througli Europe. 



HISTORY OF GEOMETRICAL METHODS 

849. Rlietorical Methods. By rhetorical methods in 
the presentation of geometric truths, is meant the use of 
definitions, axioms, theorems, geoiiieti'ic figures, the rep- 
resentation of geometi'ic magnitudes hy tlie use of letters, 
tlie arrangement of material in Books, etc. The Egyptians 
iiad none of these, their geometric knowledge being re- 
corded onlj' in the shape of the solutions of certain numeri- 
cal examples, from which the rules used must be inferred. 
Thales (Greece COO B.C.) fii'st made an enunciation of 
an abstract pro]»ei'ty of a gcometi'ie figure. He had a rude 
idea of the geometric theorem. 

Pythagoras (Italy 535, B.C) introduced formal defini- 
tions into geometry, though some of those used by him 
were not very accurate. Fof instance, his definition of a 
point is "unity having position." Pythagoras also 
arranged the leading propositions known to him in 
something like logical order. 

Hippocrates (Athens, 420 B. C.) was the tirst systemati- 
cally to denote a point by a capital letter, and a segment 
of a line by two capital tetters, as the line AB, as is done 
at present. He also wrote the first text-book on geometry. 
Plato (Athens, 380 B.C.) made definitions, axioms 
and postulates the beginning and basis of geometry. 

To Euclid (Alexandria, 280 B. C.) is due the division 
of geometry into Books, the formal enunciation of tlico- 
rems, the particular enuucialion, the f(>nual constru(;rion, 



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492 GEOMETRY. Arl'ENDrX 

proof, and cuuclusioii, Ui presenting a proposition. He 
also introduced tlie use of the corollary and scholium. 

Using these methods of presenting geometric truths, 
Euclid ■wrote a text-book of geometry iu thirteen books, 
which ■was the standard text-book on this subject for 
nearly two thousand years. 

The use of the symbols A, CO , \\ , etc., in geometric 
proofs originated in the United States iu recent years. 



850. Logical Methods. The Egyptians used no formal 
methods of proof. They probably obtained their few crude 
geometric processes as the result of experiment. 

The Hindoos also used no formal proof. One of their 
writers on geometry merely states a theorem, draws a 
figure, and says "Behold I " 

The use of logical methods of geometric proof is due to 
tiie Greeks. The early Greek geometricians used experi- 
viental methods at times, in order to obtain geometric 
truths. For instance, they determined that the angles at 
the base of an isosceles triangle are equal, by folding half 
of the triangle over on the altitude as an axis and observ- 
ing that the angles mentioned coincided as a fact, but 
without showing that they must coincide. 

Psrthagoras (i325 B. C.) was the first to establish geo- 
metric truths by systematic deduetion, but his methods 
were sometimes faulty. For instance, he believed that the 
converse of a proposition is necessarily true. 

Hippocrates (420 B. C.) used correct and rigorous de- 
duciion iu geometric proofs. He also introduced specific 
varieties of such deduction, such as the method of reduc- 
ing one proposition to another (Art, 296) , and the reductio 
ad ahsurdtim. 



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HISTORY OF GSOMETKY 493 

The methoda o£ deduction used by the Greeks, ho^vaver, were de- 
fective in their lack of generality. For instance, it was often thought 
necessary to have a separate proof of a theorem for each different 
kind of figure to which the theorem applied. 
Thus, the theorem that the sura of the an- y-~~~~7\ 

gles of a triangle equals two right angles ,-' 'v, /' '\ 

(1) for the equilateral triangle Tiv use '•■ / .. . . v 
of the regular hexagon ; 

(2) for the right triangle by the use of 
a rectangle; 

(3) tor a Boalene triangle by dividing 
the scalene triangle into two right triangles. 

The Greeks appeared to fear that a 
general proof might be vitiated if it were 
applied to a figure in any way special or 

peculiar. In other words, they had no conception of the principle o( 
eontinuity (Art. 843). 

Plato (380 B. C,} introduced the metliod oE proof by 
ait<'!ysi'^. that is, by taliing a proposition as true and work- 
ing from it back to known truths (see Art. 196) . 

To Eudozus (3G0B. C.) is virtually due proof by the 
method of limits; though his inetliod, known as the 
method of exhaustions, is crude and cumbersome. 

ApoUonius (Alexandria, 225 B. C.) used projections, 
transversals, etc., which, in modern times, have developed 
into the subject of projective geometry. 

851. Mechanical Methods, Tlie Greeks, in demonstra- 
ting a geometrical theorem, usually drew the figure em- 
ployed in a bed of sand. This method had certain advan- 
tages, but was not adapted to the use of a large audience. 

At the time when geometry was being developed in Greece, the 
interest in the subject was very general. There was scarcely a town 
but had its lectures on the subject. The news of the discovery ol a 



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4',)i GEOJIETKY. .VPI'EyDIX 

new theortm spreati from town to town, and the theorem nas redenjon- 
strated in the sanii of each raarket place. 

The (ireek treatises, however, were written on velhim 
or papyrus by tiie use of the reed, or calamus, and ink. 

In Roman times, and in the middle ages, geometrienl 
figures were drawn in wax smeared on wooden boards, 
called tablets. They were drawn by the nse of the stylus, 
a metal stieii, pointed at one end for making marks, and 
broad at the other for erasing marks. Tliese was tablets 
were stiil in nse in Shakespeare's time (see Hamlet Act 1, 
Sc. 5, 1, 107). The blackboard and crayon are modern 
inventions, their nse having developed within the last one 
hundred years. 

The Greeks invented many kinds of drawing instruments 
for tracing various curves. It was due to the inflneuee of 
Plato (1180 B. C.) that, in constructing geometric figures, 
the use of oniy the rnier and compasses is permitted. 



HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY, 

852. Rectilinear Figures. The Egyp- 
tians measured the area of any fonr- 
sided field by multiplying half the sura 
of one pair of opposite sides by half the 
sum of the other pair; which was equivalent to using the 

lormuia, area — — ^X -n ~' 

This, of ooursB, gives a eorreet result loc the rectangle and square, 
but gives too great a result for otlier quadrilaterals, as the trapeaoid, 
etc. Hence Joseph, of the Book of Genesis, in buying the fields of 
the Egyptians for Pharoah in time of famine by the use of this 
(ormuia, ill many onses paid for a larger field than lie obtained* 

The Egj-ptians had a special fondness for geometrical 
constructions, probably growing out of their work as temple 



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HISTORY OF GEOMETRY 495 

biiilders. A ekss of workers existed among them called 
"rope-stretchers," whose business was the marking out of 
the foundations of buildings. These meu knew how to 
bisect an angle and also to coustruct a right angle. The 
latter was probably done bj' a method essentially the same 
as forming a right triangle whose sides are three, four and 
five units of length. Ahmes, in his treatise, has various 
constructions of the isosceles trapezoid from different data, 

Thales (600 B, C.) enunciated the following theorems: 

If two straight lines intersect, the opposite or vertical 
angles are equal; 

The angles &i the base of an isosceles triangle are equal; 

Two triangles are equal if two sides and the included 
angle of one are equal to two sides and the included angle 
of the other; 

The sum of the angles of a triangle equals two right 
angles; 

Two mutuallj" equiangular triangles are similar. 

Thales used the last of these tiieorems to measure the 
height of the great pyramid by measuring the length of 
the shadow cast by the pyramid and a!so measuring the 
length of the shadow of a post of known height at the same 
time and making a proportion between these quantities. 

Pythagoras (525 B. C.) and his followers discovered 
correct formulas for the areas of the principal rectilinear 
figures, and also discovered the theorems that the areas of 
similar polygons are as the squares of their homologous 
sides, and that the square on the hypotenuse of a right 
triangle equals the sum of the squares on the other two 
sides. The latter is called the Pythagorean theorem. 
They also discovered how to constnict a square equivalent 
to a given parallelogram, and to divide a given line in 
mean and extreme ratio, 



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496 GEOMETRY. Ari'ENDIX 

To EudoxuB (380 8. C.) we owe the general theory of 
proportion in geonietr.v, and the treatment of incommen- 
surable quantities by the method of Eshaustions. By the 
use of these he obtained such theorems as that the areas 
of two circles are to each other as the squares of their 
radii, or of their diameters. 

In the writings of Hero (Alexandria, 125 B. C.) we first 
find the formula for the area of a triangle in terms of its 
sides, K=Vs(s — a) (s — b) (s — e). Hero also was the first 
to place land-snrveyiug on a scientific basis. 

It is a curious fact tliat Haro at the same time gives an ineorreet 
formula for the area of a triangle, ¥iz., K=ia{i+':), tliis formula 
being apparent!}' derived from Egyptian sources. 

Xenodorus (150 B. C.) investigated isoperemetricai 
figures. 

The Romans, though they excelled in engineering, ap- 
parently did not appreciate the value of the Greek geom- 
etry. Even after they became acquainted with it, they 
continued to use antiquated and inaccurate formulas for 
areas, some of them of obscure origin. Thus, they used 
the Egyptian formula for the area of a quadrilateral, 

K=—^ X— ^. They determined the area of an equilat- 
eral triangle whose side is a, by different formulas, all 
incorrect, as K=~~- , K=i{a"+a), and ZL = ia^. 



853. The Circle. Thales enunciated the theorem that 
every diameter bisects a circle, and proved the theorem 
that an angle inscribed in a semicircle is a right angle. 

To Hippocrates (420 B. C.) is due the discovery of 
nearly all the other principal properties of the circle givaa 
in this book. 



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HISTORi" OP GEOMETRY 497 

The Egyptians i-pgarded the area of the circle an eL|iii\ni- 
lent to ,ff of the diameter squared, which would make 
T-3.iC04. 

The Jews and Babylonians treated 7t as equal to 3. 

Archimedes, by the use of inscribed and circumscvihed 
regular polygons, showed that the true value of 7t lies 
between Sf and 3ki; that is, between 3.14285 and 3.1408. 

The Hindoo writers assign various values to 7t, as 3, 3b, 
l/lO, and Aryabhatta (530 A. D.) gives the correct ap- 
proximation, 3.1416. The Hindoos iised the foi'mnla 

1/2 1/4 A& '^^^ ■^''*- *^^^ ^^ computing the numeri- 

eal value of ti. 

Within recent times, the value of n has been computed 
to 707 decimal places. 

The use of the symhoi n for the ratio of the circum- 
ference of a circle to the diameter was established in 
mathematiea by Euler (Germany, 1750). 



HISTORY OF GEOMETRIC TEDTHS, SOLID GEOMETRY 

854. Polyhedrons. The Egyptians computed the vol- 
umes of solid figures from the linear dimensions of such 
figures. Thus, Ahmes computes the contents of an Egyp- 
tian bani by methods which are equivalent to the use 



is not known, it is not possible to say whetlier this formula 
is correct or not. 

Pjrthagoras discovered, or knew, all the regular poly- 
hedrons except the dodecahedron. These polj-hedrons were 
supposed to have various mugica! or mystical properties. 
Heuce the study of them was made very iiromiaeut, 



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49S GKOMKTRV. APl'ESinX 

Hippasus {470 B.C.) discovered the dodecaliodrnn. but, 
he was drowned by the other Pythagoreans for boasting 
of the discovery. 

Eudoxus (380 E. C.) showed that the volume of a pyra- 
mid is equivalent to one-third the product of its base by 
its altitude. 

E. F. August (Germany, 1849) introduced the prisma- 
toid formula into geometry and showed its importance. 

855. The Three Round Bodies. Eudoxus showed that 
the volume of a cone is equivalent to one-third the area of 
its base by its altitude. 

Archimedes discovered the fonuulas for the surface and 
volume of the sphere. 

Menelaus (100 A. J).) treated of tlie properties of 
spherical triangles. 

Gerard (Holland, 1620) invented polar triangles and 
found the formulas for the area of a spherical triangle and 
of a spherical polygon. 

856. Noa-Euclideaa Geometry. The idea that a space 
might exist having different properties from those which 
•we regard as belonging to the space in which we live, has 
occurred to different thiuiers at different times, but 
Lobatchewsky (Russia, 1793-1856) was the first to make 
systematic use of this principle. He found that if, instead 
of taking Geom, Ax. 2 as true, we suppose that through a 
given point in a plane several straight lines may be drawn 
parallel to a given line, the result is not a series of absur- 
dities or a general reductio ad absurdum; but, on the con- 
trary, a consistent series of theorems is obtained giving 
the properties of u space. 



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ITT. REVIEW EXERCISES 

EXERCISES. CROUP se 

REVIEW EXERCISES IN PLAXE GEOMETEY 

Ex. 1. It the bisectors of two adjacent angles are parpendioulaj 
to each otber, the angles ai'e supplementary. 

Ex. 2. If a diiigonal of a quadrilateral bisects two o£ its angles, 
the diagonal bisects the quadrilateral. 

Ex. 3. Through a given point draw a secant at a given distance 
■om the center of a given circle. 

Ex. 4. The bisector of one angle of a triangle and of an exterior 
ngle at another vertex form an angle which is equal to one-half thtj 
lird angle of the triangle. 

Ex. 5. The side ol a square is 18 in. Find the ulrcumference of 
the inscribed and eiroumseribed circles. 

Ex. 6. The quftdrilateral jlDiJC iK inscriHeii in a circle. The diag- 
o.iiils Af! and iJCintersect in the point F. Arc ^J> = 112°, Rrc AC = 
lAFC=^ li". Find all the other angles of the figure. 

Ex. 7. Find the locus of the center of a circle which touehee two 
given equal circles. 

Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm.,, 
210 em. 

Ex. 9. The line joining Ihe midpoints of two radii is porpendiculaf 
to the line bisecting their angle. 

Ex. 10. If a quadrilateral be inscribed in a circle and its diag- 
onals drawn, bow many pairs of Bimilar triangles are focnied t 

Ex. 11. Prove that the sum of the exterior angles of a polygon 
(Art. 172) equals four right angles, by the use of a ligure formed by 
drawing lines from a point within a polygon to Uie vortices of the 
polygon. 

(430) 



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GE05IETEY. 



Ex. 12. In a circlo whose radius is 12 em., find the langth of the 
lugent drawn from a point at a distance 240 mm. from the center, 

Ex. 13. If two sides of a regular pentagon he prodiieed, End the 



Ex. 14. In the parallelogrEm ABCD, points are taken on the 
diasoaals such that AP=BQ=CE==DS. Show that PQBS is a 
parailelogram. 

Ex. IB. A chord in. long is at the distance 4 in. from the eentcv 
of a circle. Find the distance from the center of a aiiord 8 in. long. 

Ex. 16. If B is a point in the eireumferenee o£ a eicele whosa 
center is 0, PA a tangent at any point P, mooting OB producBd at .-), 
and PD perpendicular to OH, then. PB bisects the angle AFD. 

Ex. 17. Construet a parallelogram, given a side, an angle, and a 
diagonal. 

Ex. 18. Find in inches the sides of an isosceles right triangle 
whose area is 1 sq. yd. 



Ex. 20. 3f two lines intersect so that the product of the segments 
of one line equals the product of the segments of the other, a cir- 
cumference may he passed through the extremities of the two lines. 

Ex. 21. Find the locus of the TerticBB of all triangles on a given 
base and having a given area. 

Ex. 22. On the figure p. 206, prove that^'*+4p = ZB^H-f(^^- 

reetangle is 108 and the base is throe times 



Ex. 24. If, on the sides AG and BC o£ the triangle ABC, the 
squares, AD and BF, are constructed, AF and DB are equal. 

Ex. 25. If the angle included between a tangent and a secant is 
half a right angle, and the tangent equals the radius, the secant 



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REVIEW EXERCIHES IN PLANE GEOMETRY 501 

Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the 
difference of their areas is 15 sq. yds. Find their radii. 

Ex. 27, Construct an isosceles trapeaold, given the bases and s, 



Ex. 28. Show thfit, i£ the alternate sides of a regular pentagon 
he produced to meet, the points of intersection formed are the -vertices 
of another regular pentagon. 

Ex. 29. If a poat 2 ft, fi in. high caats a shadow 1 ft, 9 in. long, 
liow tall is a tree which, at the same time, casts a shadow H6 ft. long I" 

Ex, 30. If two intersecting chords make equal angles with the 
diameter through their point of intersectiju, the chords are equal. 



Ex. 31. From a gireii point dra 
einal segment Is half the secant. 



rele whioh touehus 



Ex, 33. If one diagonal of a quadrilateral bisects the other 
diagonai, the firat diagonal divides the quadrilateral into two equi- 

Ex. 34. In a given square inscribe a square having a given side. 

Ex. 35. A fleld in the shape of an equilateral triangle contains 
one acre. How toaiij feet does one side contain f 

Ex. 36. IE perpendiculars are drawn to a given line from the ver- 
tices of a parallelogram, the sum of the perpendiculars from two 
opposite vevtiees equals the sum of the other two perpendiculars. 

Ex, 37. Any two altitudes of a triangle are reolproeally propor- 
tional to tbe bases on which they stand. 

Ex. 38. Construct a triangle equivalent to a given tj-iangle and 
having two given aides. 

Ex. 39. The apothem of a regular hexagon is 20, Find the area 
ot the inscribed and circumseribBd circles. 

Ex. 40, M is the midpoint of the hypotenase A£ of e, right tri- 
angle ABC. Prove 8 ilC~ = jJs"+llC~+Tc''-, 



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50'-i C.KOMF/l'llY. ArrHNlHX 

Ex. 41. Traiififonn a civon tvianjjlo into an e 
angle eontaiuijig a given acute angle. 

Ex. 42. The area of a square inscribed in a 
aiea of the square inscribed ia the oirole as 2 : 'i. 

Ex. 43. If, on a diameter of the oirele 0, OA- 
aliel to BD, the chord CD is perpendicular to JC, 



Bi. 4-5. State and prova the converse of Prop. XXI, Bonk III. 

Ejt. 46. If, in a giren trapezoid, one base is three times the otlifc 
base, the segments of each diagonal are as 1 : .t. 

Ex. 47. If two Bides of a triangle are 6 and 12 and the anglt- 
included by them is liil°, find the length of tlie other side. Also lind 
this when the iuciMdod angle ia 4."!° ; also, when ll!0^. 

Ex. 48. How many aides has a polygon in whieh the sum of the 
interior angles exceeds the siita of the exterior angles by 540"? 

Ex. 49. If the four sides of a quadrilateral are the diameter of a 
oirele, the tvo tangents at its extremities, and a tangent at any other 
point, the area of tlie quadrilateral equals one-half the pcoduet of the 
diameter by the side opposite it in tlie quadrila-teral. 

Ex, 50. An equilateral triangle and a regular he.iragon have the 
Bsrae perimeter; End the ratio of their areas. 

Ex. 51. To a circle whose radius is 30 e 
from a point 21 dm. from the c 

Ex. 52. If two opposite sides ot a quadrilateral are equal, and 
the angles which they make with a third side are equal, the quad- 
rilateral is a trapezoid. 

Ex. 53. If two circles are tangent ostornally and two parallel 
fliametera are drawn, one in each circle, a pair of opposite extremities 
of the two diameters aud the point of contact are eolllnear. 

Ex. 54. If, iu the triangle ABC. the line AD is perpendicular to 
BD. the bisector of the angle II, a line through V parallel to BC 
bisects AC. 



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EKVIFAV EXERCISES IN PLANE GEOMXTRY i)0-i 

Ex. 55. BLaocl a given triangle by a line parallel to a given liae. 

Ei. 56. It two parallelograms have au angle of one e-^uul to the 
Eupplement ot an angla of the other, their areas ai'o to each other aa 
the produetB of the sidoa inoluding the angles. 

Ex. 57, The sum of the medians of a triangle is less than the 
perimeter, and greater than half the perimeter. 

EiC. 58. If PARIS is a secant to a <>,irele through the center O, FT 
a tangent, and Tit perpendicular to Pli, then I'.l ; l'lt = I'0 : PIS. 

Ex. 59. Two concentric circles have radii of 17 and 15. Piud the 
length of the ehord ot the larger which la taugtrnt to the smaller. 

Ex.60. Onthe egure, p, 244, 

(o) Find two pairs of similar triangles; 

(6) Find two dotted lines whieli are pei'pendiculac to each other; 

(e) Diaeover a theorem oout-erning points, not connected by lines 
on the figure, which are eollinear; 

{(J) Discover a theorem concerning sqnares on given lines. 

Ex. 61. One of the legs, AC, of an isosceles triangle is produced 
through the vertex, C, to the point F, and F is joined with I), the mid- 
point of the base AB. I)!'' intersects IlC in £. Prove that Cf ia 
greater than CE, 

Es. 62. The line of centers of two circles intersects their common 
external tangent at P. PAliCI) is a secant intersecting one of the 
two circles at A and U and the other at C and D. Prove PA X PD = 
PEXPC. 

Ex. 63. Trisect a given parallelogram by lines drawn throut'h a 
given vertex, 

Ex. 64. Find the area of a triangle the sides of which are tho 
chord of an arc of 120° in a circle whose radius is 1 ; the chord of an 
arc of UO" in a circle whose radius is 2; and the chord of an arc of 
60° in a circle whose radius ia 3. 



Ex. 66. Two circles intersect at P and V. The chord CQ is t, 
gent to the circle QPIl at Q. APB is auy ehord through /'. Pn 
that AC is parallel to Bq, 



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Ex. 67. la tho trianglo ABC, from 7», tlio midpoint of Jli; J 
and DF are firiiwn, biseetlng tbe angles ATJIl and ADC, and maeli 
AB at E and AC nt F. I'rove £F || ISC. 

Es. 68. Produce the side KC of the triangle ABC to a point />, 
that PBXfC=I^j'. 

Ex. 69. lu a giveu circle inscribe & reetan^le similar to a f;i\ 
rectangle. 



Ei. 70. In a given sonii 


.cin'lo i 


ns.:illji) a rtcliinjjiii similiir to ?i 


given rectcngie.' 






Ex. 71. The area of an i 


soSL'oles 


trapcKoid is 140 aq. ft., one bas« 


la ^6 ft., and the legs make < 


in angle 


or 4:,- Willi tlic other liaae. Find 


the other base. 






Ex. 72. Cut off one-thli 


rd tho a 


rea of a given trinngie by i, iin^ 


perpendicular to one side. 






Ex. 73. Fiod tbe sidea 


of a tri^ 


ingl^ wLo-^o area is 1 sq. ft., if 


the sides ave in the ratio 2 : 


: :i : 4. 





Ex. 74. Divide a given line into two parts such that the aum of 
the squares of the two parts shall be a minimum. 

Ex, 75. If, from any point in tbe base of a triangle, lines are 
drawn parallel to tbe sides, find the loeus of the center of tbe paral- 
lelogram ao formed. 

Ea. 76. Three siiios of 'a quadrilateral are 845, 613, filO. and the 
fourth side is pei'pendieular to the siijes H45 and HIO. Find the area. 

Ex. 77. It BP bisects the angle ABC, and BP 
bisects the angle CHA, prove that angle F=i sum 
of angles A and C. 

Ex. 78. Two circles intersect at F and Q. 
Through a poiut A m one oireumferenoe lines Al'C 
and AQD are drawn, meeting the other in C and 1). Pr 
gent at A parallel ta CD. 

Ex. 79. In a given triangle, draw a line parallel to tbe base and 
terminated by tho aides so that it shall bu a mean jn'oportional be- 




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REVIEW EXEIiCISES IN PLANE GEOMETRy 505 

Ex 80. Find the angle iasoribed in a asmioirole the Bum of whose 
rides is a maximum, 

Ex. 81. The bases of a trapezoid are IGO and 120, and the alti- 
tude liO. Find the dimensions of two equiTalent trapeioids into 
which the given trapezoid is divided by a line parallel to the base. 

Ex. 82. If the diameter of a given circle be divided into any two 
segraente, and a seraioireumfereQce be described on the two segments 
on opposite aides of the diameter, tlie ai'ea of tlie civcle will bo di- 
vided by the semieii'cumferencea thus drawn into two pai'ts having 
tlie same ratio as the segments of the diameter. 

Ex. 33. On a given straight line, AB, two segments of oiroiea are 
drawn, ^PJSnml AQH, Tlieanglea QAI' tmiQBF are bisected by Unas 
meeting in E, Prove that the auglo £ is a constant, wherever P aiiJ 
Q may be on their arcs. 

Et. 84. On the side AB of the triangle ABC, aa diameter, a cir- 
cle is described. AT is a diameter parallel to BC. Show that EH 
bisects the angle ABC. 

Ex, 85. Construct a trapezoid, given the bases, one diagonal, and 
an anfjlo included by the diagonals. 

Ex. 86. It, through any point in the common chord of two inter- 
aectinf; circles, two obonls be drawn, one in each circle, through the 
four extremitiee of the two chords a ciruumference may bo passed. 

Ex. 87. Prom a given point as center dsserihe a circle cutting a 
given straight line in two points, so that the produot of the distances 
of the points from a given point in the line may equal the square of a 
given line segment. 

Ex. 88. AB is any chord in a given circle, P any point on the 
cireumference, P\[ is perpendicular to AB and is produced to meet 
Iha circle at Q; AN is drawn perpendieular to the tangent at P. 
Prove the triangles A'AM and PAQ similar. 

Ex. 89. If two circles ABCD and EBCF intersect in B and a and 
have common exterior tangents AE and D-F aut by BC produced at <j 
audi/, thi^u VH'- = Ba--\-AE-. 



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aUh GKOMKTKY. Al'fENiilX 

EXERCISES. CROUP BT 

RKVIi;\V EXERC'lfiES IN SOl.lD liEOMETRY 

Ex. 1. A sepment of a straiRht line oblique to a piane is greatei- 
than its projeclioti on Ihe plane. 

Ex. 2. Two tetvahedroiiri are aimilar if a dihedral angle of oim 
equals a diliedral angle of the other, and the faeea forming tboso 
difieilral angles are similar each to each. 

Ex. 3. A plane and a straight line, both of wliic'h are parallel tc. 
the same line, are parallel to each other. 

Ex. ,4. It tlie diagonal of one taue of a cube is 10 iurlits, find Iht, 
volume of the cube. 



Ex. 5. Construct a spherical triangle o 


■n a ijiv 


lea of the sides of the triangle. 




Ex. 6. Given AJ! 1 MX, 




Jfand KF I. MR; 




prove EF X I'M. 




Ex. 7, The diagonals of a reetangula 


r paral- 




Ex. 8. What portion of the surface of a sphere 
is a triangle eseli of whose angles is 140"? 

Ex. 9. Through a given point pass a plane parallel to two given 
straight lines. 

Ex. 10. Show that the lateral area of a cylinder of revolution is 
equivalent to a circle whose radius is a mean proportional between 
the altitude of the cylinder and the diameter ol its base. 

Ex. 11. The volumes of polyhedrons circumscribed about equal 
spheres are to each other as the surfaces of the polyhedrons. 

Ex. 12. Find Sand T of a regular square pyramid sn edge of 
whose base is 14 dm., and whose lateral edge is 250 cm. 

Ex. 13, It two lines are parallel and a plane be passed through each 
hese pianes is parallel to the given lines. 



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EEVIEW EXERCISES IN SOLID GEOMETRY 



Ex. 14. Givoii Pll X plane AD, 
Z rEU= Z FFH ; 
prove lI'El'=Z.rFE. 



El. 15. If a plniie be passed 
through the midpoiiila of three 
edges of a pnrallelopiped wliiirh 

meet at a vertex, the pyramid thna formed 13 what purt of the 
parallelopiped f 

a of its dist.anp«s 



Es. 17. Given the points A, Jl, C, I> in a plane and Pa point 
outside the plane, J/1 perpendieuldr \o the phme PBD, and AC p.T- 
pendiKular to tlie plane i'VIi; prove lliiit I'D Is perpeudieuhir tu 
the plane Al'.CIi. 

Ex. 18. In n spherb whoso radius is ">, find the aifii of a /nnti tiju 
radii of whose upper and lon-er hases are -T and 4. 

Ex. 19. Two cylinders of revolution havK equal lateral aroaK, 
Show that their volumes are as It : It'. 

Ex. 20. The midpoints o£ two opposite sides ot a quadrilateral in 
space, and the midpoints of its diagonals, are tlie vertioes ot pi 
parallelogram. 

Ex. 21. How many feet oE two-ineh planlt are necessary to eon- 
Btrucf. a box twice as wide as deep and twice as long aa wide (ou the 
inside}, and to contain 216 on. ft.! 

Ex. 22. If two spheres with radii K and r are concentrie, find the 
nreu of the section of the larjjer sphei'e made by a plane tangent to 
the smaller sphere. 

Ex. 23. In the frustum of a regular square pyramid, the (!.!;.■■■» 
of the bases are denoted by b\ and hi and the altitude by H ; pvusu 



that X=iv'(6i— i-il'-h^fl^ 



Ex. 24. If the opposite slJ 
V .>|..posite angles are equal. 



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508 (jEOiii'iTKY . APPi;>; dix 

Ex. 25. Obtiiin the simplest formiita for the lateral suvfaee of a 
truDeated tvianfrular right prnm, each edge of whosf base is «, and 
whose lateral edges are 21, 9, and r. 

Ex. 26. The area of a zone of one ba^o is a mean proportional 
between the reiuainiiig surface of the sphere and its eiitiio surface. 
Find the altitude of the zone. 

Ex. 27. The lateral edges of two similar frusta are as 1 ; <i. How 
do tiieir areas compare ? Tlieir volumes f 

Ex. 28. Construct a spherical surface with a given radius, r, which 
shall be tangent to a given plane, and to a given sphere, and also pasa 
through a given point. 

Ex, 29. The volume of a right circular cylinder equals the area 
of the generatiuf- rectangle multiplied by the ciroumferenee generated 
by the point of intersection of its diagonals. 

Ex. 30. On a sphere whose radius Is Si inches, find the area of a 
zone generated by a pair of compasses whose points are 5 inches 

Ex. 31. The perpendicular to a given plane from the point where 
tlie altitudes of a regular tetrahedron intersect equals one-fourth tho 
sum of the perpendiculars from the veitieee of the tetrahedron to the 
same plane. 

Ex, 32. Two trihedral angles are equal or syminetrieal if their 
corresponding dihedral angles are equal. 

Ex. 33. On a sphere whose radius is a, a zone has equal banes 
and the sum of the bases equals the area of the zone. Fldd the alti- 
tude of the zone. 

edges of a tetrahedron 

Ex. 35. Find the loeus of all points in space which have their 
istances from two given parallel lines in a given ratio. 

Ex, 36. It a, b, c are the sides of a spherical triangle, a', 6', c* 
iie sides of its polar triangle, and a>b>c, then a'<l/<e'. 

Ex. 37. A cone of revolution has a lateral area of 4 sq, yd. and 
n altitude of 2 ft. How much of tho altitude must be cut oft by a 
ilane parallel to the base, in order to leave a frustum whose lateral 



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EEVIEW EXERCISES IN SOLID GEOMETKY 



Ex. 38. The total a 
isoribed sphere as 9 : ' 



Ex. 39. CoQstrucl a sphere of give 
be thngent to three given, spheres. 



eu. in. and its altitude ia 20 in. Show how to find the radii of the baaes. 

Ex. 41. On each base of a cylinder of revolution a cone is placed, 
with its vertex at the center of the opposite base. Find the radius of 
the circle of intersection of the two oonieai surfaces, 

Ex. 42. The volume of a frustum of a cone of revolution equals 
the sum of q cylinder Bud a eone of the same altitude as the frustum, 
and with radii which are respectiveiy the half sum and the half differ- 
ence of the radii of the frustum. 

Ex. 43. A square whose aide is a revolves about a line through 
(mo of its vertices and parallel to a diagonal, as axis ; find the surface 
and volume generated. 

Ex. 44. If a cone of revolution, roll on another fixed cone of revo- 
lution so that their vertices coincide, find the kind of surface gen- 
erated by the axis of the roiling eone. 

Ex. 45. An equilateral triangle whose side is a revolvea about au 
altitude as au axis; find the surface and volume generated by the 
inscribed circle, and also by the eireumseribed circle, 

Ex. 46. Find the locus of the center of a sphere which la tan- 
gent to three given planes. 

Ex. 47. If an equilateral triangle whose side is a be rotated about 
a line through one vertex and parallel to the opposite side, as an axis, 
find the surface and volume generated. 

Ex. 48. What other formulas of solid geometry may be regarded 
as special eases of the formula for the volume of a prismatoid ? 

Ex. 49. Through a given point pass a plme which shall bisect the 
volume of a given tetrahedron. 



Ex. BO. In an equilateral < 
leuts are porpetidicuhir at tl 



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PRACTICAL APPLICATlOnS OF PLANE GEOMETRY 



EXERCISES. GROUP S8 

(Book I) 
1 Take 1 piece of pipir li urns i '^IiiirI 
o iaim I !i[;lit aiit,k \\ hjt gtomi I 




( liGO inil fol 1 thp paper 
I |iriiKi]ik en definition 



Wh^t RComHru 


.i.nuplK 


hu 


^ou 


nU)cout'iii!eanp:l 
halt the anfrle, j: 
thp dijgruni (tL 


..f thP c 
llPt«Cl 

note the 


■upp 
the 


iteis 
out 
bye 


l^e It IS csMutull^ the mrfhod wfd 
pe ind hcn(u m corn ctijig inslni 



itiaiuhL pcli,!, test the j,ipuiii\ of the 

I iiippJifLi^ -jqiiri ljy u meth'xl 

he tliiigr nil Ilo« thtn would 

ipLUi in of tin 111-kU uirIp of 



3 InE\ iproiethit lh( PI 
square if there he auj cqii il« o 
side lines of the square as sho«n ii 
and show that e + r = j;) 

T-his prmeiple la irapoitant btea 
in correcting the a\is of i teloioc 
menta of which thp t Icicopt h a part ; 
HUrieving and astronomical instruments 

4 By use of a ciipenters square am 
sltaight edge lay off i strits of pnnllel lines \\hat 
property of pii lilel line^ ba^e joii used' 

5 Tell how to (,onBtru.ct a carpcntoi i miter box 

6 Ihe (ii^run hhows a di miiiK instr imuit edled 
thepirill 1 lulir, lh( dotud outlm 
of thi. in-.tiuni(nt in another jiosilion th pirt RS 
leimining faxiil Thp dist mci i'Q = hS, PR = Q'^ 
HeiiLL show thit PQ ii patalltl to R!> 

In like manner show thU PQ i» imralld to RS 
Hi nee Awv. PQ i^ panllLl to P Q 

If \ line b( ilriwn p( rp< nditular to PQ and 
anothrt p(ip<ndicuUr to P Q the pctpcndiiul ir ll u~ d i 
pmllci to lach i.thir { \it 111 lin,-, i Liptu hi. In to i , ill Uii i. 
art, parjlkl) 




510 



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PHACTECAL APPLIPATIOXS f)!! 

The above are ctbcs of linked motion, a kind it mMhanism o[ mdo 
importance Observe fonnatance the system of hnks which oonntLt 
the driving wheels ot j loLomotivi nith the [iiaton in the tyhnder and 
also the jointed rods conntctiug thi walking bi im ot j, atearaboa! nith 
the engine Look up ilso, m the Cmliuv DictionavVi for instuncL 
the words hnkagt, icii, and jiaialkl motion 

7. The distdnec- bpifliiii 
paflaable bainer (h» b 
a pond) may be Idi i 
no instrument for in i 

Ut A and B Ix I hi > 
iMit Bliitioii ( a,nd meiaiirL 
to F, making ( I = l< 
DC = liC Mtasiirt tJl 

1! AC = Cr -220 ft,; liC - 
long is AB? 

8. In the trusses of steel bridges, why are the beams nnd rods 
arrai^ed so as to form a network of triangles as far as possible, and not 
ot quadrilaterala, or pentagons, for instance? (See Ex. 3, p. 70; also 
Art. 101.) 

How 13 this principle also made useotin forming the frame of a wooden 
hiiX'^Q oiboxcir or tostren^thni i « eak f i ime < r f ener of an\ kind' 
9 Draw a map for the lollowmg sar\(j Hot's to the seilo ot 
400 ft to lh( inch In Hjitir! 
off the an^k'i dia" i. dott(.d 
lorth and south line through 
a<!i stitioti ml then use a 



l^^o iCHs-ii>li ph(e, 


sepaiated bi in 


m tlioi 

1 luml 

L l( iiadHI I'lodi 
I'rodULG m U. I> 1 


;;,:■;. .^^. 

' " / \ 

,1 1,11 u. 


Phul lis = 1)1 




liC = Cl> -liiOtL; 


ami IJF - 21(1 fl., 1 



Stations 


|„, 


iiing=i 


Di,l mceh 


1 


JN 


"0 i 


m> fl 


1> 


' \ 




J)i) fl 




1 s 


r"} 


JiOfi 


' D 

1 


■^ 


.o°\\ 


bOOlt 



Kttp your dialing £or a 
iltr iiw 
10 Obtain oi nnkeupawLt 
ot Bur\e\ nolcf, snnd ir lo those in E\ md niakt a lii using tor them 
U IitD( and/Tfp ol2l be tno naih iRiiMndiiiilar to the ]>Iiti( 
fthtpipti Lit smiUmnrorslji' itluhi-d to those » Ulh at land/) 
Ul Z( - W I,,turi\ of hght pi«i Ihiough y 1o 1 bircflul I 
I 1 /i inclthimoto/' ProvctSiil/ l/'B is i right ingle 

[->LC The Ian of reflicted light is that the ai^le of inLideni e e(]n i\- 
thi angit of reflection or in the figure x = x and u = y ili(.u Jit 
thi triinuln ABC j; + ,j + i^" - ISO'' or i + y = n-i° \l out thi 
p, iitB 1 ani b it i-'t +21/ + 1 = 100° u + d = tJO ttc ] 



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(";eometry 




Bht c 



gf 



t s Kbt Rl 1 



II h 1 113 
t gl thth 
ray coming from Q through /' to A. 

IS, The velocity of light is dolermined by the 
' " in till! following manner: 
J_ pI.TJie of the paper ami 
livot; let OP, bel,-l,fi;. 
it and reflected Ihroujih 
ills distant, whence it i> 
retuni of fhe ray to 0, 




a rotating 
Let A,Bi be a mirror 
rotating about as u ] 
Let to be a ray of light striking the miritir : 
M to a small Rtafionnry mirror some m 
ref)i:ttcd back through .1/ to 0. On the 
A,B, will have rotated through a 
small angle, o, to the position AiB-; 
hence the ray will be reflected in the 
direction OR, Let the pupi! show that 
Za = I^Z LOR. Since Z. WR may 
readily be measured, ^ a is known, and 
if the rate at which the mirror is rotat- 
ing is known, the time occupied by the 
ray in traveling from to the stationary 
[Suo, ^PiOPj = Z /l.O^! (Art. 132) 

Z_P,OL = Z. PvOM = cc ( / incidence = Zreflcfliuti) 
.■- Z P^OL = x~a 
.■. ZiOfi =» + a- (x-a) =2al 
If Z LOR = 2° 19', OM = 3 mi., and the mirror AB rotates 100 
times per second, determine the velocityof light per second. 

13. To prove that the image of a point in a plane mirror is on a 
perpendicular from the point to the mirror and as far behind t!io 
mirror as the object is in front of it, let 
MM' be the mirror and P the point, and 
./> PAE and PA'E' two reflected rays. Show 



and back Js determined. 




Z PAM = AEAW = £ P'AM. 
:. Z /M.U' = Z P'AM'. 
. PAA' = 4 P'AA'. 
.-. PA = P'.i, etc. 
14. From the above show the direction in which a billiard ball must 
be sent in order to strike a certain point on the table afler striking oat 
side uf the table. 



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p 
h 




f 


\ 


7 / 


b 



/ 


-' 




PRAmCAL VPPLI'^ATION'* ■51'i 

lors 0.1/ anil 01/ bp peiprnd :hr to pt h (thpr 
makp i pon'if ruction to show 
nhere the point P mtil appear 
to be to an e>e at E aftir tbe 
light from P hia hppn iifltctpd 
from both mirrors 

hoc Pio\eSP'' = A 1 + 4P 
= B 1 + IP ] 

16 ■\\hat woull liL thL ippU- 
i ition of F^ 15 to a ball -truck 
jn a biUiaid tabU 

17 The path of a ni} of light 
before and iiftcr enlcnnji; a Rlaia 
prism IS gnen b\ ihc Imes AB 

and CD The entire angle bj «hich a 
raj of light IS defieetpd on passing thro igh 
the prism is denoted by x Prove that 
J- = , + r - P (Zi T'Bn and PC^aie 1 1 ZO 
ta. By use of d quadrilateral BPCy, 
y = 180° - P. Then use the quadrilateral 
all of whose sides are dotted lines,] 



EXERCISES. GROUP i 




2, By use of the c.irpenfer's sqiiaro find the 
center of a given circle. 

3. Show how a. pattern maker by the use of a 
carpenter's square can determine whether the 
ciivity made in the edge of a boatd or piece of 
metal is a Beniicirclo, 

4. Biboct a (civcn angle by the use of a carjientpr's sqiiaic. 

fi. By use of squared paper divide a line U inches long into five 
equal parts. Info 7 equai parts. Into 3 equal parts. Can you make 
this division on paper ruled in only one direction? 

6. Make tip ami work a similar exatjiple tor yoursolf. 



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(GEOMETRY 




long and let be ifK midpoitiL Murk off 
four equal parts at D, F, li. With O 
through C, Z), P, H. A. From draw 
radii (one of which is OA) dividing the 
angular space about into eight equal 
pxrts Beginning at dran tht 
smooth curve ( VHi-tb through thp 
joints where the ladii drawn inter 
ect the circumferpin.es ui, indieatcil 
m the diagrim Ihc lesult will be 
the outline of i special foim of wheel 
(ailed a C'jm much UM'd in machine 



t mo- 



th g aid 

] ea p t b Pe 
g 1 a; 

1 l g hoes A ; 
1 t t 

back and forth motion. TI i ff f 1 g h bet 

on the diagi'am ia termed tl ti f th 

8. Make a lirawing of } h th thi h 
the longest radius IJ in. 

9. Stretch a 100 ft. tape or part of it so ai 

10. To extend a straight line AB beyond an obslatic, ;is a building, 
we may proceed as follows: 

At B measure off an angle C ,. 

ABC = 60°. Produce CB to D, /\ /\ \ 

etc. Let. the pupil complete the a  — '— ^^ /^ — ^ — —K 

construction and prove lliat his \ / 

method is correct. V 

11. Let the pupil solve the 

problem of Ex. 10 by the construction of right angles instead of angles 
of 60°. 

12. To find the distance between two points A and B, one of which 

A l"1 ^■^^ '^ inaccessible, we may proceed as fol- 

p _ u--_j,B idYjg: Extend BA to C. Measure a con- 

Q/^''%\ venient line AD and extend AD to E, 

^^^-^y f| making DB = ^D. From £ run a line 1| 

.F^ E \\. ■^'' ^""^ meeting BD extended at F. 

^ Measure £f, Prove that £/■ = A£. 



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PRArXICAL APPLICATIONS 



515 





if' 



13 Show how to solve the problpni of F\ 12 iij' c-onsfriieting a 
hne at right angles to xnc Iher hne instead of one parallel to inotlier, 

14 If two Eirfetfi meet as in the diagram 
show hoB, a cur\o of gi\en radius r may be 
made to tike the ylaee of the angle 4 and be • 
tangent with the curb of the two btveets 

16 The curve in a railroad track is usually 
ao arc of a circle tangent to each straight track 

T which it joinn If two straight fra^-l « 4B and 

r>. '"D oto joined bj a circuUr curve tangent to 

I ofh Dt them and P la the point where IS and 
( D \o !J mft it e\tKnded piove ZTI D = 2 
Z. Tii 

16. One na> of Ujmg off a railroad curve 
n track is as follows; Let ABP be the ^ 

given straight track. At B construct a small angle PBC 
(whose size depends on the degree of curvature which 
the curve is to havet. On it mark off BC = 100 ft. y 

Construct Z CBD = Z P^C. Take C as a center and 7 
100 ft. as a radius, and describe an arc cutting BD / 
at D. Construct E from D in like manner. Prove / 
that, B, C, D, B all lie on 
/ the arc of a circumference tangent 

at B. 

[Sua. Pass a circumference through B, 
C, D. Prove that B lies on this circumfer- 
^e and that ABP is tangent to it.] 
17. Let AB and CD be two straight rail- 
O road tracks connected by an arc AF of a 
circle whose center is 0, and an aic FC 
whose center is 0', the arcs having a com- 
mon tangent at F. Prove that the angle of 
n (z) of the two straight tracks, if produced, equals the sum 
of the central angles of the two tracks. 
[Sue. X ~y + z. Use Ex. 15-1 

A curve like AFC composed of two or more arcs of different radii is 
called a compound curoe. Can you suggest why a compound curve 
should lie used in connecting railroad trajjks? 

18. If Hip two arcs which compose a compound curve lie on opposite 
sides of .their common tau^i;iil, the comjiounii curve ,ia called a reoersa 



) AB 




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GEOMETRY 



Thus on the diagram, BCD 



nnecting the 
itraight roads AB and DE. 
Discover the relation between 





\% Whiiij is ii railroad Irogl' If a ciirvod track c 
track, show that the angle of the frog (i) equals the 
central angle of the curved track (o). 

20. If two tracks which curve in the same direction 
crosB each other, find the relation between the anglo . 
o[ the frog and the central angk-s of the two curved tracks. 

21. Find the same when the two curved tracks curve in opposite 
dii'ontions. 

22. Show how to locate a gas-generating plant (P) along a given 

j^ straight road AB so that the length of pipe 

/ connecting it with two towns (C and D) 

"■■v^ / shall be a minimum. 

"'-./ , (Use Ex, 24, p. 176.) 

23, Discover and state ;; 
Ex. 24, p. 17U, similar to one given in Ex. 22. 

24. If a treasure has been buried 100 ft. from a cert; 
distant from a given straight road and a path parallel t( 
how the treasure may be found. 

26. Make up and work a similar example for yourself using the prin- 
ciple of E)i. 2, p. 167, Also using that of one of Exs. 3-S, p. 168. 

26. Prove that the latitude of a place ^n the 
earth's surface equals the elevation of the pole 
(that is, on the diagram, prove Z. OE-4 = Z.PAO). 
„ Z_ 27. Given the sun's declina- 

tion (i.e. distance north or south 
of the celestial equator), show 
how to determine the latitude 
^ ^ ' of a place by measuring the zenith distance of the 

sua- Also by measuring the altitude of tb.e sun above the boiison. 



application of 



o the road, show 




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PRACTICAL APPLICATIONS 



517 



3 and 27 i 



28, How was Peary aided by the principles of Eks. 
determining whether he had arrived at the North Pole? 

29. If on April 6 (the day of the year on which Peary was at the 
North Pole) the sun was 6" 7' north of the celestial equator, how high 
above the horizon should the sun have been as observed by Peary? 
At what hour of the day was this? 

80. The sextant is used almost daily by every navigator in deter- 
mining the altitude of the sun above the horizon, and hence the latitude 
of the ship. The construction of the sextant ia ^ 

based on the principle that if a ray of l^t be 
reflected from two mirrors' in succession, the 
change in the direction of the ray of light 
equala twice the angle made by the mirrors. 
Prove this law. 

|Su(j. Let M and N be the mirrors and the 
reflected ray be SMNO. It is required to 
prove that Z.y = 2 ^x. It is to be noted 
that Z.FNM, being an exterior angle of triangle 
NS'M, equals x + i. Hence angle MNO equals \ / 
180° -2 (e -l-t). Hence in triangle MNO, \^/ 

y + ISO" - 2 (a; + t) + 2 1 = 180°, etc.] \^. 

In the sextant, angle y is the elevation of thi" 
sun above the horizon, and x can be read on the rim of the u 





1 position V 



mirror N i» fij.ed i 
M rotates 

It maj be observed thit the prin 
fjple of this example is the ^an < la 
that pro\ed m E\ 12 p 512 

31. The diagiim show? a some- 
what simple aubstitutp tor the e\ 
tant called the angle meter. ih the 
center of the arc BC and MM' is a 
fixed mirror. The instrument is 
held BO that a ray SO from the sun 
when reflected from the mirror passes 
into the eye (A) in a horizontal direc- 
tion. Show that the elevation of tliE> 

The rim BC is graduated so (hat 



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GEOMETRY 



The angle-meter may be used to 
angles. 

What is the advantage in using a mirror on 
sort? That is, why is not the positioD of the 
across a graduated arc? 

32. Draw an easement & 



liorizontul as well a; 



instrument of this 
n observed directly 




i composing the 



e (.AB) tangent to the rake cornice Bc 
and passing through a required point A. (The 
e construction is used in layiii^ out the ease- 
ments of stair rails, etc.) 
(Use Ex. 7, p. 174.) 

33. A s^mental arch is a coraijound curve 
composed of the arcs ot thriie circUs. The method 

nof constructing a segmental arch is as follows; 
Let AB be the span and CD the altitude of the re- 
quired arcli. Complete the rectangle GADC. 
Draw the diagonaJ AC. Bisect the angles 
GAC and GCA. Let the bisectors meet at E. 
Draw EH perpendicular to AC, meetii^ AB at 
JV <md CD produced at U. Make DK equal to '^' 
DN. Then show that N is the center and A'.'l 
the radius, H the center and HE the radius, 
and K the center and KB the radius for the a 
arch. (See Hanstein's Constructive Drawing.) 

[ScG. At E draw a hne perpendicular to EH. Then Z. -VB-1 - 
Z. NAE. (Complements of equal angles are equal) etc.] 

34. What is called a Persian arch may be constructed as follows; 
Let AB be the span and CD the altitude of the required arch. Draw 
p n P *he isosceles triangles ADB. Divide AD into three 
equal parts at H and G. Construct HK ± AG 
at H, and meeting AB produced at K. Produce 
KG to meet Ef which has been drawn through D]| 
AB. With K as & center and KA as a radius, and 
. B as a center and EG as a radius, describe arcs 
 meeting at G. Prove that these area have a com- 
mon tangent at 0, and therefore form a compound 
5urve. (See Hanstein's Constructive Drawing.) 
3B. Construct a Persian arch in which the arc DG = arc NG. 
[Soa. Bisect line AD instead of trisecting it.] 
36, Construct Persian arches in whicli llio chord:; NG and DG have 




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PRACTICAL APPLICATIONS 519 

various ratios, and decide wliich of these arches you think is the most 
beautiful. 

37. Construct aegmental arches of various shapes and decide which 
of these you think is the most beautiful. 

38. Construct a trefoil, given the radius of one of its circles. 
[SoG, Construct an equilateral triangle, one of whose 

sides equals twice the given radius. Take each vertex 
of Ihe triangle as a center, and half of one of the sides 
iw a radiua and describe arcs.l f ^ 

39. Inscribe a trefoil in a given equilateral triangle, V_yV^_y 
[S0G. Bisect, the angles of the tiiu,ngle, etc.] 

40. Discover the method of eoiislrucdou of each of tlu' figuvpr! or 
diagrams on the next page, and then rcpi-oduce each <•( tiicm in a 
drawing. 

Squared paper may be used to advantage as an aid in making some 
of these constructions. See Fig. 10, p, 520. 

41. Construct a diagram siiailar to Fig. 10, but making a rectangle 
instead of e, square the basis of the drawing. 

42. .\lso one making the rhomboid the basis. 

43. Construct a trefoil and develop into !Ui ornamental design by 
placing small circles and ares within its parts and larger circles outside. 

44. By use of squared paper invent designs similar to Fig, 10 on 
p. ,')20, but formed by two intertwining lines, 

46, Collect a number of pictures of ornamental designs, tracery, 
scroll work, etc., such as arc vised in architecture, wall paper, and 
similar patterns, whose construction depends on the principles of Book 
II. Show how these designs are constructed, and reproduce them in 
drawings. 

EXERCISES. GROUP 90 

(Book III) 
1. To tind the distance between two accessible 
objcpt.-i ;1 and B separaliid by a bari'iet', take a 
convenient point C, measure AC and BC. Protluce 
EC to I) and AC to F so that DC = some frac- 
tion, as i, of CB, and FC = the same traction of 
FD. How, then, is All com- 




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rmre 





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PUACTiCAL APPLICATIONS 



521 



Let BC = 300 yd., DC = 60 yd, 
FD = 52 yd., find .IB. 

Why is this method of finding AB 
of Ex. 7, p. 5U. 

2. What similar method doea the 
gram surest for finding AB? 

3. In the di.igriim of Ex. 2, 
between A and li, and ako between 
nieaaurements would be iieeetaary 
of AB> Of BF' 

i Mork lUinCx 2 p 310 

6 In CJ.S ihi un i nnt '■hinm' 



, AC = 240 yd., FC = 4S yd., 
often more convenient than that 
adjoining dia- 





of the top tf 


1 


1 


fiom the groii 


il h 1 


ci cr tai 


the mirror l. 1 


'0 ft fi im 


the tree 


how h gh la th 


tree' 




7 I rcsters 


often detei 


mine the 


ht kH of -v tie 


e by ^n n 


stiument 


ill i raitetmms Hefei 


It Meaa- 


m IhMi 


m,,lean^ 


hich tins 


initiumtil 1 


un liiieted 


n ohoun 


in thi, di Jijr i n 






If the i -Ian 


e fiom 4 t 


) the foot 


ofthotieei bOlf h( = 


J iiichts 


mdLF =4 n 


ht■^ hi 1 


1 e 1 tight 


of the tree 






suit 


n fron -1 


roimF 



\ad shadons cannot be used the 
heisht of an objett like a 
tiet, or stteplc cin often be 
dctLrmined bj a method in 
dicated m the d r i w i n g 
W hat di'itanee mutt be meas 

lied and whv to determine 
thi, hught of the tret* 

6 'ihow how to find the 
h „ht of I tret by placing 

I 1 irior m a honzontal posi- 
tion on the ground and stand- 
ing fco as to see the reflection 

if the oijserver's eye is 5i ft. 



1 t 



, the 







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522 



r.EOMKTIiV 



ACB. On paper make adiaiving of llio triangle ACB to a convenient 
scale. On lliis drawing mraanrp l.iic Wm. wliifh represents AB and 
hence determine the Icnj^lli of AH. 

By use of this method ivc arr> riiivrd Ihn labor of marking out and 
measuring the lines CD, CF, and DF. 

Apply this rtiethod to the raeasurcraenfc of two objects in your noi^i;h- 
borhood which are aepai'ated by an impassable barrier. 

In like manner show how the distance from a given point to another 
inaecessible place may be dptermined. 

9. Also the distance between two place." both of which are inacces- 

10. What is the meanuring inst.rument called the diagonal Boale? 
How is the principle of similar trianglca tised in it? Show how tile 
diagonal seale aiils in the accurate moasurement of lines and hence in 
the accurate determination of a distance such as AB in Es. 7, p. 511. 

n 11, Id the triangle OAB, 

A is a right angle, and 0.1 
is 1. By a method which 
is Ijeyond the scope of this 
book, the length of AB is 
computed and found to be 
.839 +. Using this tact find 
the second triangle, 
of the accompanying table, find RQ if angle P is 10°. 20", 

Tables ^ving the other sides of all possible right triangles when one 
side is unity have been computed and when used as in Exe. 11 and 12, 
form the basis of the subjwt of Trigo- 
nometry, ity use of this science, after 
measuring the length of a single lino a 
few miles long on the earth's surface, we 
an d t m*n th i' ( n es and laf 
po9 1 ons f th pi i-p^ th usands f 
mis ana w ihout a. g an 
t en ng 1 n By u f tl e« esuU 
as a bas th d tan e f th moon 
dete m aed aa ipp o\in t t 40 000 
miles of the ) SOOOOO n le- 

and of th n -e t fa d s( ''0 000 

000 Omi 01)0 I 




Anglo 


AB 


08 


10 


,176 


I.OIS 





.364 


1.064 





.Till 


1.155 


40 


.839 


1.305 


)0 


1,192 


1,556 


()0 


1.732 


2.000 


70 


2.747 


2.924 





5.671 


5.759 



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PR\rTIC\L APPLICATIONS 

A knowledge of these diotances ha^ led to importinl impioi.enii 
in method'5 of iixMg,ition and havi, thus fanihtilt^ travel ^nd 
merce and increased their benefi''! foe ( 

usaU 

13 On thpdiagramgi\enAJSif,i)Onoii 
and the angles as indicated, compute the 
length of CD, by use of the table given in 
Ex. 13, 

14 To the diagram in Ex 13 annex 
another tnangle CFD giving it angles %vhi h 
are multiples of 10° and compute ll 
length of CF 

16 A yanloqraph is an in trument fui 
drawing 1 phne figun iinuhr to i rimii 
plane fagurt It is uaetl for enlai^mg oi -^ uuwu. i> 

reducing maps and dra^vmgs It consist-, of four bars, parallel in 
pairs and jomted iX c b C and E as shown in the diagram, <i)EC 
IS a parallelogram The rods may be joined so that any required 





CB' 



fixed pivot and pencils are 
at b ind F. 

Show that A, b, and B are 

equals the given ratio 



A turns upon a 
carried 



[Sua, Dra^v a line from .4. to b, and a line from A to B. Prove the 
triangle Acb and ACB similar {Art lt^3); henw show that Ab and AB 
coincide, etc.] 

16 Using the fact that a triangle whose sides are 3 4, and ^ units of 
length is a right triangle show how by Btretchmg a 100-ft tape, U> 
ponRtruet i right angle is accuiately is posi^ibte (\mong the ancient 
Egyptians a class of workmen existed called rope stretchers, whose 
busine** was to construct right angleb in this genera! w i\ ) 

17 The sttongest beam which can be cut from j 
gnen round log is found as follows Take AB s 
diameter of ihi log and trisect it at C and D Diaw 
C E and Dt ± \b n\ meeting the iucumf<rimt \\ 
h a.tidFrp«puli\(K Draw AF FB BL and If 

Pro\e iFBF i rcdinnU also FB U - 1 v^ 
or iprroxinitih i > 7 




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(324 GEOMETRY 

|SuQ. _£B is a mean proponional beUveen DB and AB. Therefore 
FB' - i AB" (Art^43).] _ 

In like mnniiM F-V - I AB', ote,] 

18. A sphere S weighing JOO ib. rests on ihe inclined plane AB. AC 
contains 8 units uf length, BC 6 units. A 
force which prevents S from rolling down 
the plane would be equivalent to a lifting 
force of how many pounds exerted on S and 
pai'ailei with l/i'' 

[Suo. Ile=!ohe the n eight of S (represented 

by SP) into two lorces, one perpendicular to 

AB and the othei parallel to .-IB. Prove the 

triangles ACB and "il'B similar, and obtain 

AB: BC = SP: SR, or 10: 6 = 100 lb SR] 

The principle involved in this example is of great practical impor- 
tance. Thus in many maoliines useful results are often obtained by 
representing a force by the diagonal of a rectangle (or parallelogram) 
and separating this force into two component forct* represented by 
the sides of the rectangle (or parallelogram), onlj one of these com- 
ponents being effective. This principle makes poa>ib!e (he action of the 
propeller of an aeroplane or steamboat, of the best wat«r wheels and 
wind mills, and indeed of all turbine wheels. It also determines the 
lifting power of the planes of an aeroplane. 

18. A wagon weighing 1800 lb. stands on the side of a hiU which has 
a rise of IS ft. for every 100 ft. taken horizontally. Hence what force 
must a horse exert to keep such a wagon from running down hill, fric- 
tion being neglected? 

20. Make up and work a similar example for yourself. 

21. Show how the diameter of the earth may be determined by the 
following method: Drive three stakes in a level piece of gi'ound (or in a 
shallow piece of water) in line, each two successive stakes being a mile 
apart, and lot each stake project the same distance above the ground 
(or water). By use of a leveling instrument, determine the amount 
by which the middle stake projects above a horizontal line connecting 
the tops of the end stakes. This distance will be found to be 8 in. 

[Sue. Use Art. 343. .\n arc a mile long on the earth's surface may 
be taken as equal lo its chord. Then from the diagram of Art. 343 we 
obtain the following proportion, 

the diameter nf the earth; 1 mi. = 1 mi.; 8 in.] 

22. AI?o show that the liistance that the middle stake projects above 



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PRACTICAL APPLICATIONS 



525 



the horizontal liiip connecting the top of the tno end stakes varies as 
the squai'e of the dista,nce between the end stakes. Thus if the two end 
stakes were placed three times as tar apart !\s in Ex. 21 [that is, 6 miles 
apart instead of 2 miles) the bulge of the eai'th between them would 
be 3' or 9 times what it was originallj-. 

Hence determine the projection of the middle stake (or bulge of the 
earth) when the end .sljikos are 4 mi. apart. Also 8 mi. 16 mi, 33 mi. 

23, At the seashore an observer whose eye was 10 ft above sea level 
observed a distant steamboat whose hull was hidden for a height of 
12 ft. above water level by the bulge of the earth. About, how far off 
Wiis the steamboat? 

24. A seaman in a lookout 42 feet above water level with a glass 
could barely see the topsail of a, distant ship, and estimated this top- 
sail to be 45 ft. a,bovs sea level. Estimate the distance of the observed 
vessel from the seaman. 

2E. Work again Exs. T-S, p. 310, Group 56. 

36. The moon's distance from the earth's center approximately 
equals 60 times the earth's radius. A body failing at the earth's sur- 
face goes 193 in. in 1 sec. Hence, if the law of 
1, the distance fhe moon falls 



gravitation is 

;. toward the earth will be 



193 ir 



Inth. 



diagram, let be the earth's center, ABE the 
moon's orbit, and CB or AD the distance thi- 
moon falls toward the earth iu 1 sec. Taking 
the month as 27 da, 7 hr. 43 min. 11 sec, 



show that .-ID 



193 i n 
60= 



approximately. This is the ealculation used 




by Sir Isaac Newton in testing the truth of the law of gravitation, 

27, Any rectangular object, as a book, door, or photograph, is con- 
sidered to be of the most artistic shape when its length and breadth 
have the same ratio as the segments of a line divided in mean and 
extreme ratio (Art. 370). In accordance with this rule, if a window is 
6 feet high, how wide should it be? 

The division of a line in extreroe and mea 
Golden Section. Byraany theGotdenSectio: 
tal principle of esthetics, having applicationi 
tions of the ideal humas form, in esplanatio 



n ratio has been termed the 
n is regarded as a f undamen- 
! in determining the propor- 
n of musical har 



. Make up and work nn example similar to Ek, 27. 



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GEOMETRY 
EXERCISES. GROUP 91 
(Book iV) 
The supporting power of a wooden beam (of rectangular cross 
n and of given length) varies as the area of the 
section multiplied by the heigiit of the beam. 
I If the cross section of a given beam is 4° X 8°, com- 
e the supporting power of the beiim when it rests 
on the narrow edge (4") with il« supporting power 
when it regis on its wide edge (8'). 

2. Two beiims of the same length and material have cross sections 
uhich are 2" X 4' and 3° X 8' respectively. Find the ratio of the 
supporting power of the two beams, 

3. In Ex. 17, p. 533, compare the supporting power of a beam cut 
from a log in the manner indicated, with that of a square beam cut 
from the same log. 

4. Also with that of a beam whose width equals i of the diameter. 

5. When an irregular area Uke ABCD i 
calculated by means of equidistant offsets, the 
following rule is used; To the halt sum of the __ 
initial and final offsets add the sum of all the S 
intermediate offsets, and muUiply tiie sum by the common distance 
between the offsets. Prove tiiis rule. 

6. Show how the rule of the preceding prob- 
, lem could be used to calculate an area whose 
re boundary is an irregular curved line. 
. Surveyors often determine y 
the area of a piece of land, as of ABCD, by taking i,' - 
an auxiliary line as NS, measuring the perpendicular 
disUnces from A, B, C, D, E to NS, and the inter- 
cepts on NS between these perpendiculars, and ^" 
combining the areas of the various trapezoids (or j 
triangles) formed. Supply probable numbers for the 
lengths of lines on the diagram, and compute the ""\ E 

area of ABODE. ^ 

, a d (, (t.. a T ^' '^f^Quently (as when the center of 

'~^~j~~~J:  i ~ '^e curve cannot be seen from the curve)' 
' K'~'--i,^^ a railroad curve is laid out by construe- 

^^'^ ting a aeries of equidistant offsets per- 
peadicular to the tangent of the curve. 



mmm 



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'■---.1 ^ 


"ir 


~--4^ 


^-~-^fi! 




■/ " 


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PRACTECAL APPLICATIONS 527 

Thus, if ABT is a straight track and BC is a Rurve to be laid out tan- 
gent tfl AB at B, marli off 6ai, aioi, ajoj, all = d, and construct the _L 
offsets a b\ a«ft" Ojfej by Uiing the forniiiU a bu ^ r — Vi^ — n'd' 
n here!" IS the radiua of the curve Prove thit this formula is correct. 

9 The diagram represent'" tno atiaight parallel railroad tracks 
connected bj a compound curve (in this ^ ^ 

caKe called a crosa-over tta<k) comiostd of p -r*\ 

t«o tracks with common tangents at D and / 1 

b with centers and and having equal -i / ; 

ridii Taking the magnitudes as indicated ) /' ]F 

oil the figure show that A/ODisareif 
angle (use 4rts 122 IfiOl 

10 Find u formula tor i m teims of ( 
d and w (TJae the right tnjnt,U PO. 
ind 4rt 401)) \lso tsr ( in tetnib of J^ 

11. If i« = 4' Hi", d - 10 ft., )■ = 1,^0 ft., i/ 
find I. '' 

12. IE a steamboat is traveling at the rate of 12 miles an hour, and a 
boy walks across her deck at right angles to her line of motion at the 
rate of 3 miles an hour, draw a diagram to show the direction ot his 
resultant motion. From this determine the spi;ed at. which he is going. 

13. Make and work a similar example for yourself concerning a mail 
bag thrown from a train. 

14. Also concerning a breeze blowing into a window of a moving 
trolley eai'. 

16. Two forces, one of 300 !b., the other of 400 lb., ai-( at right angles 
on the same body. Find their resultant. 

16. Make up and work a similar example for yourself. 

17. A river is flomng at a rat« of 4.25 mi. iier hour, and a man is 
rowing at right angles with the current at a rate of 3.75 mi. per hour. 
WTiat is the result-ant velocity of the man? 

18. If a star has a velocity of 15 mi. a second toward the earth and a 
velocity of 20 mi. a second at right angles with a line drawn from the 
star to the earth, find the velocity of the star in its own path. 

19. A body is moving in the straight line. ID past the point (p. 528). 
The distance passed over in a unit of time = AB = BC = CD. When 
the body reaches B it is acted on by a force which impels it towaj:d 



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GEOMETRY 



a diBtancc BF in a unit of 1 ii 
!■ bv ;i line draw 
by til 




Prijvi! lluit 'iio aren swopl over in t!-. 
m the body tu will be iinchangc ; 
of thf now force. (Thai is, on ilu' 
.liagram prove that AOCB =0 AOPB.) 

This prinoipie iiccoimts for the fact that the eaiii! 
(or another planet or a ooiiiet) moves fuKter in i;, 
orbit the neoi-ei' il is l-o the Kim. 

20. By traeing tho oiiliine of a map on squanvL 
paper, show hoiv to find the aj'pa of an irregular 
figure as of some oountry or pari of a country. By 
iiBU of this method find the area of aomc part of 
the state in which you live. 

exercises. group 92 
(Book Y) 

1. k cooper in fitting a head fo a barrel takes a pair of compasses 
and then adjusts them till, when applied six times in succession in the 
chine, they will exactly complete fhe eireumferonco. Hr- Ihen takes 
the distance between the points of the compasses as Ihe radius of the 
head of the barrel. Why is this? 

2. Draw a square and convert it into a regidar octagon by cutting 
off the corners. 

ISuG. Draw the diagonals of the square, bisect their angles of inter- 
section, etc.] 

3. In heating a house by a hot-air furnace the area of the cross 
section of the cold-air box should equal the sum o£ the areas of the pipes 
conducting hot air from the furnace. If a given furnace has three hot- 
air pipes, each 6' in diameter, and one pipe S" in diameter, and the 
width of the coid-air box is 1.5°, how deep should the box be? 

i What is the most conMinif nt Tva\ of determining the diameter of 
a hot air pipe if you haic no alhpers and the ends of the pipe are not 
accessible' 

6 ^ half mde running tiack is to hait equal semicit cuUr ends and 
parallel straight sides The extreme length of the rectangle together 
with the semicircular end*! is to be 1000 fl Find the width of the 
rectangle 

[Sue Denote the length of the i \d\u> ot ihe emiciriukr ends bj 
I and that of one of the parallel idi. straight traj.k=i b-v y and oLtnn 
a pair of ■■laiultaDeoua iquafioni | 



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PRACTICAL APrLlCATlONfi 529 

6, A belt runs over two wheels one ot which has a diameter of 3 ft. 
and the other of 6 in. If the first wheel is making 120 revolutions per 
minute, how many is the second wheel making. How many revolu- 
tions must the first wheel make in order that the second may make 
300 revolutions per mtrmte? 

7. If in laying a track a rail 10 ft. long is bent through an angle ot 
5° 10', what is the radius of the curve? 

Assuming {what is not strictly true, owing to friction against the 
sides of the pipe, etc.) that the rate of flow through it cylindrical pipe is 
proportional to its area of cross section: 

8, Work again Ex. 12, p. 277, 

9. If a Ij-ii- i'ipG is replaced by a 1-in. pipe, how much i,-- the flow 
of water increased? 

10. A 3-in. pipe is to be replaced by one which will deliver twice as 
mui;h water per minute. Find, to the nearest quarter of an inch, the 
diameter of the new pipe. 

11. A 2-in. steam pipe conveying steam from the boiler to the radia- 
tors in a school building is found to supply only two thirds the needed 
amount of steam. What is the smallest even size of pipe that will 
convey the needed amount? 

12. A city of 40,000 people is barely supplied with water by 12-in. 
mains from Ike reservoirs. If these mains are lorn out, and IS-in. 
mains substituted, what future population of the city is allowed for? 

13. A steel bar I in, m diameter will hold up 50,000 lb. What load 
would a bar I in. in diameter carry? What would be the diameter of a 
round (cylindrical) bar to carry 150,000 lb.? 

14. It Iho center of symmetry of a flat, homogeneous object is the 
center of mass, find (he center of mass of a square; of a rectangle; 
regular hexagon; circle. 

15. It is evident that if the medium of a triangle (BM) be placed on 
a knife edge the triangle will balance (for if P!" 
be II AC, the pull on P is balanced by the pull on 
P'). Hence find the Renter of mass for any triangle. 
Tor a regular pentagon. 

It is tiscfui to be able Id determine i.he center of 
mas-s of an object by neomi^try or by any other 
meaus, sincii a knowledge of the center of mas.s ot a Ixidy often 
pi\!il)lis us to treat the body in a simple way, tor example, as il llie 
l.o(l> iviTc concentrated at a single point. 



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5^0 nEOMKTHY 

16. if a box has a triaaguiar end, subject to thi 
points, at. u'liat single poict oa the end must a 
be applied? 

17. The cross section of a cylinder is a circle. The woiglit-supporting 
strength of a horizontaJ cylindrical beam of given material and length 
varies as the area of the cross section times its radius. 

Compare the weight-supporting power of two solid horizontal iron 
cylindrical beams of the same length and quality of iron, the radii 
being 3 in. and 6 in. respectively. (PoinI out and use the short way 
of getting the desired result.) 

IS. The cross section of a hollow cylinder (i.e. of a tube) is a circular 
ring. Denote the outside radius of the tube by R and the inside radius 
by r. Then it may be shown that the weight-supporting power of a 
hollow cylindrical tube of given length and material varies as the area 

. , . iP + r= 
of the cross section (ve. of thermg) times — -- — . 

It R = i in, and r = 3 in., compare the weighl-wiipporiinir powf^r 
of the tube, with that of a solid cylindrical beam of the same length 
and same cross sectional area. 

19. Make up and work a similar example. 

In general a cylindrical tube is stronger than a solid cylindrical beam 
of the same length and containing the same amount of material. 

Hence in a framework, as in that of an jurship where the maximum 
strength must be obttuned from a given amount of material, the metallic 
rods and posts iire all tubular. In like manner, bamboo rods, since 
theyai'e hollow, are used in an aeroplane instead of solid wooden rods 
wherever possible. For the same reasoB, the bones of flying birds, and 
many bones in men and animals are hollow and not solid 

n 20 To coni^trucf a square which '.hail be ap- 

proMm'itely equivalent to a given circle divide 
the radiun OA into four equj,! parts and produce 
each end of two perpendicular diameters a ell's 
; equal to one fouilh of the ra liiis and 
rjnnect the extremities of the lines thus formed 
slio« that taking the squaie thus foimed aa 
equn ilent t the circle i= th( sime as taking 
" = 3J, Also find th( per <u!t of erroi in tikmg thi quare aa 
equivalent to the circle 

21. A short wiy to construct a regular inscribed pentjg n ^.n I aho 
il five-pointed stii l w | Liitagrarr ) i-. aa followa Dii \ a ii le and 




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PUACTICAL APPLICATIONS 



531 



two diamelors /IB atiti CD at right angles. Bisect the radius OB at F, 
and with f as a center and FC as a radius describe an arc cutting AO 
at H. Then CIl is the length of a side of 
the regular inscribed pentagon, by joining 
whose alternate vertices the pentagram 
niay be formed. 

As a proof of this, by use ot right tri- 
angles, prove 

pVio- 2V-5 



CH - fl— 



C^ee E\ 1" p 301 ) 





22 f irp iitpra some 
lies use the follow mg ^ 

thod of approximately deteimining the length 
t a circumference whose ladiu'i is Inown: Let 
O be the giien circle Dran Oi ind OB, radii 
t right angles Draw AB j,n1 the radius OELAB 
,t D Measure DE Then take cncuraference 
^ =b iO + DE 

Find the per cent of erroi in this method, taking t = 3 14159-. 
23. The following designs are important in architecture or in orna- 
mental work (thus the first is a detail in a stained glass window in an 
early French cathedral; the second is the plan of the base of a coS- 
umn in an English cathedral). Discover a method of constructing each 
of tiic following designs, and reproduce each of them in a drawing: 




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532 CEOMETRY 

21. Construct a square. Take each vertex of the square as a ccater 
and one half a side of the square as a radius and describe arcs which 
meet. Erase the squai'e and you have a quatrefoi!. By drawing 
other circles ajid arcs of circles, elaborate the quatrefoil into an orna- 
mental design (see design 11, p. 520). 

26. In like manner consli'uct a cinquetoil by use of a regular penta- 
gon and develop it into an ornamental design. 

26 Troj,t I icgular JicMgon m the same way. 

27 4 pa^tmeiif or raouiL may be formed out of regular polygons in 
thf, folloivmg '^^^b Shou how to m:i.kc each diagram in the simplest 




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APPLICATIONS OF SOLID GEOMETRY TO MECHANICS 
AND ENGINEERING 



EXERCISES. GROUP 93 



(Books \'I and VII) 



a the fl:it 



1 A carpenti 
edge to the surface in various diriK;t; 
the flitne^s of a will surface? Wha 
these mechanics' 

2 EJipHin nhj in object with I 



of a surface by applying a ; (might 
w does a plasterer test. 
ic principle is used by 




I'eii legs, as a stool or tripod, 
object with tour legs, iis a table, 

ever use foiar-leggcd pieces .of 

furniture? 

3. How can a earpt;nter 
get a comer post of a house 
in a vertical position by use of 
a carpenter's Btiuaro? What 
geometrical prineiple does he 

4. The diagram is the plan 
ot a hip roof. The slope of 
each face of the roof is 30°. 
Find ihe ienglh of a hip rafter 



sAB. 



1 (he pli 



trianRle DBC repre,wnli 
Hence it may l>e fih 



El section ot the roof at 
20 
I ISC = - v;!. In 



like manner by taking a section through 
BF, it is found that AC = 10. Hence in 
%e triangle ABC, AB may be found.] 

6. Find the area of the entire roof 
represented in Ex, 4, 

6. Make drawing showing at- wiiat 
fttigle Ihe two ends of a rafter like liC in Ex. 4 




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5^4 GEOMETRY 

7. Make drawings ?liowinR at what angle a jack raftpr like 12 in 
Ex. 4 rauBt be cut. 

ISuiJ. To determine how the end 1 of the jaek rafter must be eut 
use the principle that two intersecting straight fines determine a plane 
(Art. 501). The cutting plane at 1 must make an angle ;it ihe side 
of the jack rafter equal to angle CBH, and on the top of the jiick rafter 
equal to angle ABC] 

8. What is a gambrtjl root? Make up a set of examples coticorning 
a gambrel roof similar to Exs. 4-7. 

9. By use of Art. 615, show that a page of this book held at twice 
the distance of another page from the same lamp rei'ei\'es one fourth 
the light the first page receives. 

10. The supporting power of a wooden beam varies directly as the 
area of the erosa Heotion times the height of the beam and inversely as 
the length of the beam. Compare the supporting power of a beam 
X2 ft. long, 3 in. wide, and 6 in. high with that of a. beam 18 ft. long, 4 in, 
wide, and 10 in. high. -Uso compare the volumes of the two beams. 



EXERCISES. GROUP 94 

(Books VIII and IX) 

1. A hollow cylinder whose inside diameter is 6 in, ia partly filled 
with water. An irregularly shaped piece of ore when placed in the 
water causes the top surface of the water to rise 3.4 in, in the cylinder. 
Find the volume of the ore. 

2. What is a tubular boiler? What is the advantage in using a 
tubular boiler as compared with a plain cylindrical boiler? If a tubular 
boiler is IS ft. long and contains 32 tubes each 3 in. in diameter, how 
much more heating surface has it than a plain cylindrical boiler of the 
same length and 36 in. in diameter? (Indicate both the long method and 
the short method of making this computation and use the short method.) 

3. If a bridge is to have its linear dimensions lOOO times as 
great as those of a given model, the bridge will be how many times 
as heavy as the model? 

Why, then, may a bridge be planned so that in the model it will sup- 
port relatively heavy weights, yet when constructed according to the 
model, falls to pieces of its own weight? 

Show that this principle applies to other ronstructions, such as 
buildings, machines, etc., as well as to bridges. 



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rRACTICAL APPLICATIONS 535 

i. Work again Eks, 23-25, 27, p. 473. 

6. Make and work for yourself an example similar to Ek. 24, p. 473. 

6. SouniJ spreads from a center in the form of tlie surface of an 
expanding s[)here. At the distance of 10 yd. from the source, how will 
the surface of this sphere compare with its surface as it was at 1 yd.? 
How, then, does the intensity of sound at 10 yd. from the source com- 
pare with its intensity at a distance of 1 yd.'f 

Does this law apply to all forces which radiate or act from a center 
aa to light, heat, magnetism, and gravitation? Why is it called the 
law of inverse squai'es? 

7, If a body be placed within a spherical shell, the attractive forces 
exerted upon the body by different parts of the shell will balance or 
caned each other. Hence a body inside the earth, as at, the foot of a 
mine, ia attracted effectively only by the sphere of matter whose radius 
is the distance from the center of the earth to the given body. Hence, 
prove that the weight of a body below the surface of the earth varies 
as the distance of the body from the center of the earth. 

[Stjg. If W denote the weight of the body at the surface, and w its 
weight when below, R the radius of the earth, and r the distance of the 
body from the center when below 
the surface, show that 

8 The heht ol tl e sun falling 
on a smaller sphere, as on the earth oi the moon cau-,es that b^dv to 
cast a conical shadow. Denoting the radium of the sun bj R the 
radius of the smaller sphere by r, the diatanct between the two spheres 

dr 
by d, and the length of the shadow by I, show that I =  -  _ ■. 

Find I when d - 92,800,000 mi,, R = 433,000 mi., r = 4000 mi. 
9. If in a lunar ecli|}se the moon's center should pass through the 
axis of the conical shadow, and the moon is traveling at the rate of 
2100 mi. an hour, how long would the total eclipse of the moon last? 

How long if the moon's center pas.?ed through the earth's conical 
ahadon- at n distance of 1000 mi. from the axia of tlie cone? 



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531; c i)':oMr,'ri!v 

10, If the moon's diaraotcr is 2160 mi., finil Ihn length o( the moon's 
shadow us caused by sunliglit. 

11. If the distance of the moon from the earth's eenter VMrics from 
221,600 mi. to 232,970, show bow this explains why some ei;lipscs of the 
mm are total and others annular. 

Why, also, at a given point on the earth's surface is an eclipse of the 
sun a so much rarer sight than an eclipse of the moon? Why, also, is its 
duration so much briefer? 



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FOEMi.n.AM OF" PLAICE OEOMETEY 

SYMBOLS 

6, <!=siiles of Uianglo J/tr. /,'--^iadiH3 of a tilrcle. 

s = i{a + b + c). i).-diameter of acirele. 

ftf = aUitude on sklo r. (' = cirpiimferenee of acirele. 

5ii( = meiliiin on side ('. f = i'iitlius of an inso'lbed 

*c=biaettorotanglB.>i.!"Hile pirole. 

SLiii.' .-. T = -r-apprax. (or :i. 141(3— ). 

id)n=liiiB segnjouLs. K=area. 

/■= [loi'iuietcr. /i = liase of a triangle. 

6\i = slde of a regular polyi^oii h =altitude of a triangle, 

ol » sides. ''[ anil ?i- = bases of a trapezoid. 

LENGTHS OF LINES 

1. In a right triangle, C being the tight angle, 

f' = n' + &'. Art. 346. 

'2. In a riglit, triangle, ( and m being Ihe projections of a and fe 

f, and li, tho altitude on c, 

3. In an oblique triangle, m being tho projectioo of b on e, 

if aisopposite an ai^^ito Z, a' = b= + c= — 2 cX m. Art. 319. 
if a is opposite an obtuse Z, a' = 6' + e' + 2 cX*"- Art. 350. 



».-iJ/<«-<.)(--»)l— «J 


Art. 393. 


me=iT/at«^ + 6"j-c^ 


Art. 353. 


If I and m are the segments of 
opposite, a:J' = l : 


Art. 3li3„ 

c made by the biaeotor o( the 
m. Arts, 333, 336. 



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ISDEX OF DEFiXlTIOXS ASD FOEJIULAS 



^bbrp It on 
ilgebr ic analy 3 
Algebra e u eti o 
■ilternat on 
Altitude of eone 

of frustum f 
of frustum of p 
of parallelogr n 
otpr.m 
of pyrom d 
of spher c I 
of trapt 
of tr ngle 
of zone 

Anil J olu o 

An{,!c 

at CD 
to 

diled 

formed hv a 

etra ght I n 
formed bv t 
inscr bed n 



I 



Angle of lull 
poljhedrnl 

right 

salient , 

sides of . 

splierieal 

straight 

tetraliedial 

trihedinl 



vertex ot 
Angles, adjacL'i 
alteniato-iiili 



llOH)ologOll;^ . 

interior , . 
of polygon 
of quadriliiti 
of spherical 



Appolom 



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IND]:X OF IH'.FINITIOXS AXD FOTiMlXAS 



major , . 

Archimedes 
Arcs, eonjligiite 
Area ol surface 
Aryabhatta 
August, E. F. 
Auxiliary linea 

Axioms, generiil , 
geometric 

Axis of circle of splie 
of circular cone 
of regular pyiaoiii; 

of symmetry 

Babylonians . 
Base of cone . 

of isosceles triai 

of pyramid 

of spherical pyr; 

of spherical sect 

of triangle , 



! of < 



lulcr 



of fnislum of or 

of frustum of ]ii 

of parallelogram 

of prism 

of spherical segment . 

of trapezoid 

of zone . . 
Bisector of triangle . 
Bodies, the Three Round 

Center of circle . 
of regular polygon 
of sphere .... 





-'■' 


Chord 

Circie 10 

are of 

center of ... . 

circumference ul". tormiil: 

for 

circumscribed . . . 

formula for area of 


103 
,406 
103 
103 
103 

274 
103 
103 


inscribed .... 
radius of ... , 
sector of ... , 

small 

Circles, conecnliic . . 
escribed .... 


105 
10.3 
104 
427 
ID.i 


rireum-centcr of Irmn^'le 
Civcumtercncp . . . IJ 
Classification of ]>islvln-dion 
of triangUM . . . 

Complement .... 
Composition . , , . 


81 
.103 
3G0 
2,33 
125 
16 
181 


Conclusion .... 
Coninirrent lines . 


24 
80 


Cone 

altitude of ... 

^^xis of 

base Of 

circular 

circular, fotiiiuhi for vo 

uine of ... . 
circular, formulas for la 

eral and total area . 


412 

412 
413 

413 

418 

417 



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INDEX OF DEFINITIONS AXD FORJIULAS 



Cone, elements of 

lateral surface of . 

oblique circular 

of revolution . 

right circular . 

vertex of 
Cones, similar 
Congruent figures 
Conical surface . 

directrix of 

element of . 

generatrix of . 

vertex of . . 
Conjugate area 
Conaequenta 
Constant . 

geometrical . 
Constants . 
Continued proportioi 
Continuity, principle of . 
Converse of a th< 
Corollary . 
Cube . . . 
Curved spaces 
Curved surface 
Cylinder , 

altitude of . 

bases of 

circular . 

circular propei 

elements of . 

lateral surface of . 

oblique 

of revolution 

of revolution, formulas fc 
lateral and total arc; 
of . . . 

right . . . 

right ciroula. 



Cylinder, right 
Cylinders of re' 
Cylindrical sut 



elor 



t of . 



D t mn d 


Pl 




D g n 1 


pol 


K 


f p lyl 


J 




f q d 


1 t 




D m t 


 


1 


t ph 






Dh d 1 


1 





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I^■DEX OF DEFIXmOXS -VXD FOIi:\IUIA 



Duality, Fi^ 



■ipk of 



Edge o£ dihedral ni, 
Edges of polyhedral 

of polyhedron . 
Egj-ptiiins  . 

, . 400, 401, 432, 
Elements of conical f. 

of cylindrical surface . 
Enunciation, general 

pftiticular 
Equal figures . 
Equivalent figures . 

solids .... 

Euclid 

Eudosua . . .493 

Euler 

European .... 
Extreme and mean ral 
Extremes .... 

Faces of dihedral angle , 

of poiyhedTiil angle 

of polylirdmn . 
Figure, curviliiiuar 

geometrical . 

mixtilinear . 

plane_ . . . 

Figures, roctilinoar 

Formulas of Piano Geo net 

for lengths of 1 ne j 

of Plane Oeoniet fo 

areas of plane fi re^ 

of Solid Geometry fc 

of Solid Geometry ft 
volumes , 
Foot o£ perpendicular 



Fourth propoilionul 
PruBtuni of toiiu. 
altitude of . , 
bases of . . . 
formula for l.ileral 

of .... 
formula for volume i 
lateral surfacp of , 
slant height of 
Frustum of pvraniid 
altitude of . . . 



base 



of 



slant height of 



;al ! 






of c.rlintirical sui 
Geometric solid , 
Geometry . 

epochs in development of 4 

history of . 

modern . 

Non -Euclidean , 



Jrigu 



of 



e ka 


400 


102, 403, 494 


Ha n on i 





. , . 108 


Hept oon 




.... 74 


He o 




. . .496 


He 1 on 




... 74 


He h i on 




. . .360 


H n loos 




400, 432, 497 


Hpp U3 




, . .408 


H ppoc atus 




491, 492, 406 


Hstor ofg 


t 


490-498 


Honol go 3 


a =1 


.... 34 



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IXDEX OF DEFINITIONS AXD FOKJIULAS 



Homology, Principle ot 


487 


I ne d ided tein^llv 


Hyperapafie .... 


48S 1 


ot centres 


Hypoteniiae .... 


33 


I rallel to pi 


Hypothesis .... 


. 24 


I rj end eul r t j 1 


Icosahedron .... 


, 3G0 


° 


In-center of triiinplE! 


. s; 


L a au 1 irv 


Inclination of line to pia 


e 347 


concurre t 




. 1-25 


Lobatche 1 


ratio 


. 1:25 


Loci^ 


Inference, immediate . 


. 23 


Log eil U t)l 


Inversion 


. 180 


Lune 


Isoperi metric figures 


. 280 


inn-lp ot 
for i f 


Jews 


, 43- 


apler al degr es 
torn nU for or a 


Lateral nrea of frustum 


ol 


square un ts ot a 


pyramid, formula for 


. 379 










ot pyramid .... 


. 37 


Alagn tude n 


Lateral edges of prism 


3G1 


mcommen 1 1 


of pyramid 


377 


Alaji nun 


Lateral faees of prism 


ill 


"Mean p opo t a 1 


of pjramid 


3" 


Mea 


Lateral surfaoe uf cnne 


41' 


Me h n cal n et} 1 


o£ ej Under 


401 


M d an of trapezo 1 


of frustum ot eon-^ 


414 


ot tranje 


Legs of iso.^d<^luaa> 


3 


"\Ienela s 


of riplit trnn„-lL 


31 


Method ilgebi 


of trapwoid 
Iinut 


(1 
\1 


of 1 ts 
Method loj, c 1 


Limits, metiiol of 


\1 


meeh i -x] 


Line 


1 


ot n I 


broken 


11 


tion« 


curbed 


n 


hcto e 1 


divided exttnmllj 


1')^ 


Metr e sjste 


dnided hatmonii ,\h 


I'lS 


^\ nor n <■ ot <■ 1 


di\ided in p\titiiic i 


ii 


■\I te 



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INDEX OF DKriMTiONS AN)") FOILMULAS 



Nappes of cono . 
Negative quaniili™ 

Non-Eup]idean gcoi 



Numerical t 



Octagon 
Octahedron 
Opposite of a thci 
Origin of geoiiK'tr 
Ortlio-eenter 



Parallei lines 
Parallelogram 

altitude of . 

bases of 
Parallelopiped 

right . . 

rectangular 
Parallel plants 
Pentagon . 
Pentedecagon . 
Perimeter of p'll 

of triangle . 
Perpendicular liii 

planes 
Pi {«) . . 
Plane . . . 

determination 
Pianes, parallel 
Plato . . , 
Point . . . 
PointB, coney lie 
Polar distance of 

triangle , , 



Poljhediai angles, equal 
synimetTical 

Polyhedron 

diagonal of 
edges of , 



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IKDEX OF DEFINITIONS AND FOT!ML"LAS 



Polyhedron 1 1 
Polyhedrons 
classificiti n 

Postulnfi. 

of solid gpon 
Postulites 

logical 
Prism 

altituclc if 



base 



of 



oirpumsLribed aboi 

inscribed in oil! 
lateral area o!  
lateral edges of 
lateral fates of 
oblique 
quailrdnjrular 

right 

Tight sectirn i 

trianguhr 

tmncuteJ 
Pri'imaioid 

bases of . 

formula for voli 
Prismoid . 
Problem 
Pi ejection of lint 

of line on plan 

of point on line 

of point on plar 
Proof 

by auperpciiiio i 

forms of 

method? 1)1 
proportion 

eontinu.d , . 

torins of . . 



third 



axi ,f 
base of 

circumscriljei about 
fiustum of 
jnbcniel in on 

latcnl ed„ ol 



lalor 



fie 



qmdr nfnil r 
regular 

regular slint 1 tif,! 
spherical 
trnngul r 
truncated 
verte\ of 
PTtha' 



491, 492, 4!)S, 4!i7 



Q d t f 

Qu d 1 t 1 

angl f 



n ! tud 
II3 p i>ortionaI 
1 p neipic of 



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IXDEX OF DEFINIT10^:S AND FORMLXAS 



Rhombus 


(it) 


f.o!id^ phTBin! 


Right section of <L hnloT 


4f)5 


Spaei-- cuned 


of pusm 


3bl 


Sphere 


romaiis 4 


4 4')0 


ftilS of 


Round liiijic Tl t n rep 


a-^s 


center of 


Eeholium 


23 


Lircuniscrihtd \bna 


Vp 1 t line 


104 


hedron 


Bfond 


17 


diameter of 


Sect 


14 


formula fot arpx 


Election o! I 1 ilJ un 


J1)0 


face of 


Sector of cink 


104 


formula for ^ohlnl 


of cirplo fori lU fjr 




great circle of 


of spilt re 


4bl 


inscribed in polili 


Sectors amiJ.r 


275 


poles of 


Segrapnt of i tlo 


114 


radma of 


of line 


14 


small circle of 


of sphere 


4b2 


Spherical angle 


Segments un ilar 


2T3 


degrees 


Semicircle 


101 




Seniicireumfeceii"e 


103 


SpheriPil poh„ua 


''ides of angle 


14 


angles of 


of poljgon 


73 


sides of 


of quadiilateril 


u6 


vertices of 


of triangk 


32 


Sphericil piramid 


Similar cones of r oluti 


on 113 


base of 


eUmdeis oi i ol li ii 


404 


vertex nf 


polvgina 


(i3 
130 


Spherical sector 
base of 


pohhedrons 


J03 


formula for lolum 


Btttors of circlps 




Spherical segment 


segments of civ(1p9 


2-j 


altitude of 


Slant height of coik 


413 




of frustum of cone 


414 


formulis tor loluu 


of frustum of pyramid 




of one bnie 


of regular pyr-iinid 


^77 


Spherical triangle 


Solid 




bi- rectangular 


geometry .... 


18,319 


formula for area 


Solids, equivalent . . 


. 303 


sqnnre unifs of 


geometric .... 


. 11 


tri-rpcfangular 



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INDEX OP DEFINITIONS AND FOEilULA; 



Spheiicil trim, 

mtntal 

Bymmetrici 1 

Spheneal wedgt. 

Spherid 

Stiaight line 
Superposition 
Supplement 

area of 
conical 



Tangent circles 

langent conimon evternil 

Ime to tpleie 

plane to cjlinJer 
plane to &phei 

Terms of a proportioi 
Tetrahedral angle 
Tetrahedron . 
Thales . , 
Theorem 
Theory of liniil 
Third proport: 
Transversal 
Trapezium 
Trapezoid . 
altitude of . 



Iriangl 


d' 7-1 


acute 


33 


altituilts jt 


33 


base of 


33 


bisectors of 


34 


bisectois of forniU 


for 213 


clagaifications ut 


3', 33 


equiangular 


33 


equilateral 


S2 


isosceles 


32 


medians of 


34 


oltuse 


33 


pohr 


441 


ii^ht 


33 


icilcne 


32 


spheueal 


438 


\crfc\ angle of 


33 


\eitex of 


33 


riihedral angle 


349 


bi rectangulir 


330 


isosceles 


350 


lectanguUr 


350 


tiirtotangular 


350 


Ungula .... 


. dlil 


Unit of measure . - 




of surface . . . 


. 231 


of volume . . . 


, . 3S3 


Units of angle . . 


, 17 


of spherical surface 


. 452 


Variable .... 


. . 120 


Vertex ani^le . 


. . 33 


Vertex of angle . . 


. . 14 


of polyhedral angle 


. , 349 


of pyramid . , 


. . 377 


of spherical pyramid 


, 461 



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IXDliX OF nEFIXITlOXS AND FORMULAS 



Vut'teic of triangle . 
Vertices of polygon . 

of polyhedron . 

of quadrilateral 

of spherical polygo 

of triangle . 
Volume of solid . 



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